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Equations of Motion Problems with Answers

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Equations of Motion Problems with Answers

This article gives you several problems and solutions related to the kinematic equations of motion. These workout questions allow the readers to test their understanding of the use of the kinematic equations of motion to solve problems involving the one-dimensional motion of objects.

In order to understand the problems and solutions, first, recap quickly the topics given below!

  • Formulas of the equations of motion.
  • How to find out the acceleration of an object?
  • What are the different types of velocity and how to calculate the velocity of a body?
  • Learn the concept and formula to find the relative velocity 

Now, you are advised to go through each problem given below and practice the strategy used to solve that question.

Problem 1:

A jeep moving along a straight highway with a speed of 115 km/h was brought to stop within a distance of 400m. What was the retardation of the jeep assuming it to be uniform and how long did it take to stop?

Answer:

large begin{aligned} We;know,;from;the;third;&equation;of;motion, \v^{2}&=u^{2}+2as \&Here;given,;v=0;; \&u=115;km/h \&=frac{115times1000}{60times 60} \&=31.94;m/s;; \&s=400;m;; \Substituting,;we;get,;0^{2}&=31.94^{2}+2atimes 400 \therefore ;a&=-left ( frac{31.94^{2}}{2times 400}right ) \&=-left ( frac{1020.16}{800} right ) \&=-1.28;m/s^{2} \therefore Retardation&=1.28;m/s^{2} \\Again,;;;;;;;;;;;;v&=u+at \therefore Time;taken,;;;;t&=frac{v-u}{a} \&=frac{0-31.94}{-1.28} \&=24.95;s \That;is,;the;jeep;will;ta&ke;24.95;seconds;to;stop. end{aligned}

Problem 2:

A man standing on a lift throws a ball upwards with the maximum initial velocity he can and is found to be equal to 55 m/s. After what time the ball returns to his hand if (a) the lift is stationary, (b) the lift is moving up with a uniform velocity of 7 m/s, (c)the lift is moving down with a velocity of 7 m/s. Also given g = 9.8 m/s2.

Solution:

Let x0   be the position co-ordinate of the hand on the man when the ball is thrown up and xt be that when the ball returns to the hand. Let t be the time taken.

large begin{aligned} Then,;;x_{t}-x_{0}&=v_{0}t+frac{1}{2}at^{2};;;;;;;-----------(1) \(a);when;the;lift;is;stationary,;&x_{t}-x_{0}=0; \&v_{0}=55;m/s; \&a=-g=-9.8;m/s^{2}; \&t=;? \Substituting;the;values;in;th&e;above;equation;(1),;we;get, \0&=55times t+frac{1}{2}times -9.8times t^{2} \0&=55t-4.9t^{2} \therefore;; t&=frac{55}{4.9} \&=11.22;s \ie,;when;the;lift;is;stationar&y;the;ball;returns;to;the;man's;hand;after;11.22;seconds. end{aligned}

(b) As the lift starts moving upwards with uniform velocity, there is no change in the relative velocity of the ball with respect to the man(ie, 55 m/s). Therefore, the ball will return to the man’s hand after 11.22 seconds.

(c) Similarly, as stated above, here also the relative velocity of the ball with respect to the man is 55 m/s and hence the ball will return to the man’s hand after 11.22 seconds.

Question 3:

The velocity acquired by an object moving with uniform acceleration is 60 m/s in 3 seconds and 120 m/s in 6 seconds. Find the initial velocity.

Answer:

large begin{aligned} We;know,;;;;;Acceleration,;a&=frac{change;in;velocity}{time} \&=frac{120-60}{6-3} \&=20;m/s^{2}. \Let;u;be;the;initial;velocity;an&d;v;be;final;velocity,;then, \v&=u+at \therefore ;;;u&=v-at \Here,;;v=60;m/s;;a=20;m/&s^{2};;t=6-3=3;s;;u=;? \therefore ;Initial;velocity,;u&=60-20times 3 \&=underline{underline{0;m/s}} end{aligned}

Problem 4:

An object moving with uniform acceleration has displacement of 25 metre is 5 seconds and 36 metre in 6 seconds. Calculate the initial velocity and acceleration.

Answer:

large begin{aligned} We;know;the;displacement,;;;;;;;;;;;;;;;;;;;s&=ut+frac{1}{2}at^{2};;;;;;;------(1) \When;t=5;sec,;the;above;eqn;(1);implies;;;;;;;;25&=utimes 5+frac{1}{2}atimes 5^{2} \implies;;;;;;;;25&=5left ( u+frac{5}{2}a right ) \ie.,;;implies;;;;;;;;;;5&=u+frac{5}{2}a;;;;;;;;;;;------(2) \When;t=6;sec,;the;eqn;(1);implies;;;;;;;;36&=utimes 6+frac{1}{2}atimes 6^{2} \implies;;;;;;;;36&=6left ( u+frac{6}{2}a right ) \ie.,;;implies;;;;;;;;;;6&=u+3a;;;;;;;;;;;------(3) \Substracting,;(3)-(2),;implies;(6-5)&=(u+3a)-left ( u+frac{5}{2}a right ) \implies;;;;;;;;;;1&=u+3a-u-frac{5}{2}a \ie.,;implies;;;;;;;;;;1&=3a-frac{5}{2}a \ie.,;implies;;;;;;;;;;1&=frac{1}{2}a \ie.,;Acceleration,;;;;;;;;;;;a&=underline{underline{2;m/s^{2}}} end{aligned}

large begin{aligned} \Now;substituting;the;value;of;a;in;eqn;(2),;;;implies;;;;;;;; 5&=u+frac{5}{2}times 2 \ie.,;;;;initial;velocity,;;u&=5-5 \&=underline{underline{0;m/s}} end{aligned}

Tutorial 5:

A bike starting from rest acquires a speed of 40 m/s in 20 seconds. Then, the bike travels with this speed for 15 seconds. Find the (a) acceleration (b) distance travelled during acceleration and (c) the total distance travelled.

Solution:

large begin{aligned} \Here,;;;;; u=0;m/s;;v=40;m/s;;&t=20;s;;a=? \\(a);We;know,;;;;;;;v&=u+at \therefore a&=frac{v-u}{t} \&=frac{40-0}{20} \&=underline{underline{2;m/s^{2}}} \\(b);We;know,;distance;travelled,;s&=ut+frac{1}{2}at^{2} \&=0times 20+frac{1}{2}times 2times 20^{2} \&=underline{underline{400;m}} \\(c);Distance;travelled;during;constant;velocity,;s_{1}&=vt \&=40times 15 \&=600;m \therefore Total;distance;travelled&=400+600 \&=underline{underline{1000;m}} end{aligned}

Kinematic Problem 6:

A hunded metre sprinter increases his speed from rest uniformly at the rate of 1 m/s2 upto 70 metre, and covers the rest 30 metres with uniform speed. How much time does he take to cover the first half and second half of the run?

Answer:

large begin{aligned} \For;the;motion;upto;70;m,v^{2}&=u^{2}+2as;;;;------(1) \Given,;u=0;m/s;;a=1;m/s^{2};;s=70;m;;v=?; \therefore ;eqn;(1);becomes,;v^{2}&=0^{2}+2times 1times 70 \&=140 \therefore v&=sqrt{140} \&=11.83;m/s \Time;taken;to;cover;70;m,;t_{1}&=frac{v-u}{a} \&=frac{11.83-0}{1} \&=11.83;s \Time;taken;to;cover;the;remaining;30;m,;t_{2}&=frac{30}{11.83} \&=2.54;s \\Let;t;be;the;time;needed;to;cover;the;first;50;m. \Then,;s&=ut+frac{1}{2}at^{2} \Here,;u=0;m/s;;a=1;m/s^{2};;s=50;m;;t=? \therefore substituting,;we;get,;50&=0times t+frac{1}{2}times 1times t^{2} \ie.,;t^{2}&=100 \therefore t&=underline{underline{10;s}} \\Total;time&=11.83+2.54=14.37;s \therefore Time;taken;to;cover;the;second;half&=14.37-10 \&=underline{underline{4.37;s}} end{aligned}

Example 7:

A ball is thrown upwards with a velocity of 55 m/s. Find the velocity after 4 seconds. Also find out the maximum height attained by the ball.

Solution:

large begin{aligned} \Velocity;v_{t};after;t;seconds,;v_{t}&=v_{0}+at \Here,;v_{0}=55;m/s;;a=-g=&-9.8;m/s^{2};;t=4;s \\At;t=4;s,;;v_{4}&=55-9.8times 4 \&=underline{underline{15.8;m/s}} \\At;the;highest;point;velocity;is&;zero, \Then,;v_{t}^{2}&=v_{0}^{2}+2a(x_{t}-x_{0}) \Here,;v_{t}=0;m/s;;x_{t}-x_{0}=h,;&the;maximum;height;attained \therefore, h&=frac{v_{t}^{2}-v_{0}^{2}}{2a} \&=frac{0-55^{2}}{-2times 9.8} \&=underline{underline{154.34;m}} end{aligned}

Question 8:

A man throws a ball upwards with an initial velocity of 34 m/s. To what height does the ball rise and after how long does the ball return to the player’s hand. Also, g = 9.8 m/s2.

Answer:

large begin{aligned} \In;the;equation,;v^{2}=u^{2}+2as,;v=0;at;the;&highest;point,;u=34;m/s,;a=-9.8;m/s^{2} \therefore ;the;height,;s&=frac{v^{2}-u^{2}}{2a} \&=frac{0-34^{2}}{2times -9.8} \&=underline{underline{58.98;m}} \\When;the;ball;returns;to;the;man's;hand,;&s=;0 \In;the;equation,;;s=ut+frac{1}{2}at^{2},;s=0,;u=34;&m/s,;a=-9.8;m/s^{2} \therefore;substituting,;;0&=34t+frac{1}{2}times -9.8times t^{2} \therefore time;taken,;t&=frac{2times 34}{9.8} \&=underline{underline{6.94;s}} end{aligned}

Tutorial 9:

A balloon is rising vertically up with a velocity of 35 m/s. A stone released from it reaches the ground after 12 seconds. Find the height at which the stone was released.

Answer:

motion related problems
Figure 1

large begin{aligned} \The;direction;of;motion;of;the;balloon;is;taken;as;&the;positive;x;direction.\Let;x_{0}=0;be;the;position;co-ordinate;of;the;ston&e;at;the;instant;of;releasing;and;x_{t};that;at;the;ground. \We;can;write;the;equation;as,;;;;;;x_{t}&=x_{0}+v_{0}t+frac{1}{2}at^{2} \Here,;x_{t};is;negative,;x_{0}=0,;v_{0}=35;&m/s,;t=12;s,;a=-9.8;m/s^{2} \therefore ;x_{t}&=0+35times 12-frac{1}{2}times 9.8times 12^{2} \&=0+420-705 \&=-285;m \ie.,;the;height;at;which;the;stone;was;released&=285;m end{aligned}

Question 10:

An object is moving along the x-direction with a constant acceleration of 8 m/s2. At time t = 0, its position is 20 m away from the origin and has a velocity of 4 m/s. Find the position and velocity at t = 3 seconds. Where does the object reach when the velocity doubles?

Answer:

large begin{aligned} \Position;of;the;particle,;;;;;;x_{t}&=x_{0}+v_{0}t+frac{1}{2}at^{2} \Here,;x_{0}=20;m,;v_{0}=4;m/s,;&a=8;m/s^{2},;t=3;m/s \therefore substituting,;we;get,;the;position;at;t=3;s,;;;;;;x_{t}&=20+4times 3+frac{1}{2}times 8times 3^{2} \&=20+12+36 \&=underline{underline{68;m}};from;the;origin \Again,;velocity,;;;v_{t}&=v_{0}+at \&=4+8times 3 \&=underline{underline{28;m/s}} \ie.,;the;position;and;velocity;of;the;object;at;t=3;seconds;&is;68;m;and;28;m/s;respectively. \\Again,;v_{t}^{2}&=v_{0}^{2}+2aleft ( x_{t}-x_{0} right ) \therefore x_{t}&=x_{0}+frac{v_{t}^{2}-v_{0}^{2}}{2a} \Here,;v_{0}=4;m/s,;v_{t}=2v_{0}=2times 4=8;&m/s,;a=8;m/s^{2},;x_{0}=20;m \therefore substituting,;we;get,;x_{t}&=20+frac{8^{2}-4^{2}}{2times 8} \&=20+frac{48}{16} \&=underline{underline{23;m}} \ie.,;the;velocity;of;the;object;doubles;when;it;reaches;23;m;&from;origin. end{aligned}

Example 11:

A ball thrown vertically upwards from the top of a building of height 80 metres returns to the earth after 10 seconds. What is the velocity of the projection? Also g may be taken as  9.8 m/s2.

Solution:

equation of motion problems
Figure 2

large begin{aligned} \When;the;ball;reaches;the;earth,;the;&displacement,;s=-80;m. \We;know,;;;;s&=v_{0}t+frac{1}{2}gt^{2} \Here,;t=10;s,;g=-9.8;m/s^{2} \Substituting;in;the;above;equation,;we;get,;;;-80&=v_{0}times 10+frac{1}{2}times -9.8times 10^{2} \implies;;;;;;;-80&=10v_{0}-490 \ie.,;the;velocity;of;projection,;;v_{0}&=frac{-80+490}{10} \&=underline{underline{41;m/s}} end{aligned}

Practice Problem 12:

A man running with constant acceleration covers a distance of 40 metres in the 4th second and 60 metres in the 6th second of its motion. What is the acceleration of the running person?

Answer:

large begin{aligned} \Distance;covered;in;the;t^{th};second,;s_{t}&=u+frac{a}{2}(2t-1) \Substituting;the;values,;we;get,;;40&=u+frac{a}{2}(2times 4-1) \implies40&=u+frac{7}{2}a;;;;;-----------(1) \Similarly,;;60&=u+frac{a}{2}(2times 6-1) \implies60&=u+frac{11}{2}a;;;;;-----------(2) \Substracting,;eqn;(2)-(1)implies;;60-40&=u-u+aleft ( frac{11}{2}-frac{7}{2} right ) \implies20&=2a \therefore the;acceleration,;a&=frac{20}{2} \&=underline{underline{10;m/s^{2}}} end{aligned}

Question 13:

Two trains X and Y are moving on parallel rails with a uniform speed of 65 km/h in the same direction with X ahead of Y. The driver of Y decides to overtake. He accelerated the train at the rate of 2 m/s2. After 2 minutes, if the two drivers could see each other face to face, what  was the original distance between them?

Answer:

large begin{aligned} \Let;l;be;the;distance;between;the;two;drivers;&when;they;were;in;uniform;motion. \we;know,;the;equation,;s&=v_{0}t+frac{1}{2}at^{2} \Here,;s=l,;v_{0}=0,;a=2;m/s^{2},;t=2;&minutes;=120;seconds \Substituting,;we;get,;;;l&=0+frac{1}{2}times 2times 120^{2} \&=underline{underline{14400;m}} end{aligned}

Problem 14:

An object falling from rest describe 75 metre in the last second of its fall. Find the height from which it fell and the total time taken to fall. Also g = 9.8 m/s2.

Answer:

large begin{aligned} \distance;travelled;in;the;t^{th};secod;of;motion,;s_{t}&=u+frac{1}{2}aleft ( 2t-1 right ) \Here,;u=0,;s_{t}=75;m,;a=9.8;m/s^{2} \Substituting,;we;get,;75&=0+frac{1}{2}times 9.8times left ( 2t-1 right ) \&=4.9(2t-1) \&=9.8t-4.9 \therefore t&=frac{79.9}{9.8} \&=underline{underline{8.15;s}} \\In;the;equation,;s&=ut+frac{1}{2}at^{2},;;;;;u=0,;t=8.15;s,;a=9.8;m/s^{2} \therefore Distance;travelled,;s&=0+frac{1}{2}times 9.8times 8.15^{2} \&=underline{underline{325.4;m}}\ie.,;the;height;from;which;it;fell;=;325.4;m;∧time;taken;to;fall;=8.15;s end{aligned}

Example 15:

An object is dropped from a height of 54 metres. What will be the distance travelled by it during the last second of its fall? Also, take g = 9.8 m/s2.

Solution:

small begin{aligned} \Let;the;total;time;of;fall;be;&t. \we;know,;s&=ut+frac{1}{2}gt^{2} \Here,;s=54;m,;u=0,;g=9.8;m/s^{2},;t=? \Substituting,;we;get,;54&=0+frac{1}{2}times 9.8times t^{2} \ie.,;t^{2}&=frac{54}{4.9} \therefore t&=3.32;seconds \\Distance;travelled;in;the;last;second;of;its;fall,;s_{t}&=v_{0}+frac{1}{2}g(2t-1) \&=0+frac{1}{2}times 9.8(2times 3.32-1) \&=underline{underline{27.64;m}} end{aligned}

Motion Problem 16:

The two ends of a train running with constant acceleation pass a certain point on the ground with velocities v1 and v2. Show that the velocity with which the middle point of the train passes the same point is small begin{aligned} sqrt{frac{v_{1}^{2}+v_{2}^{2}}{2}} end{aligned}.

Answer:

large begin{aligned} \Let;l;be;the;length;of;the;train;and;a;be;th&e;acceleration. \Then,;;v_{2}^{2}&=v_{1}^{2}+2al \therefore v_{2}^{2}-v_{1}^{2}&=2al \Let;v;be;the;velocity;of;the;train;when;its;∣point;crosses;the;given;point. \Then,;;v_{2}^{2}&=v_{1}^{2}+2afrac{l}{2} \Substituting;the;value;of;2al,;we;get,;;;;v_{2}^{2}&=v_{1}^{2}+frac{v_{2}^{2}-v_{1}^{2}}{2} \&=frac{2v_{1}^{2}+v_{2}^{2}-v_{1}^{2}}{2} \&=frac{v_{1}^{2}+v_{2}^{2}}{2} \therefore v&=sqrt{frac{v_{1}^{2}+v_{2}^{2}}{2}} end{aligned}

Question 17:

Two stones P and Q are thrown simultaneously with a speed of 30 m/s. The stone P is thrown vertically upwards while stone Q is thrown vertically downwards from a height of 90 m above P. When and where the two stones meet? Take the value of g as 9.8 m/s2.

Answer:

equation of motion practice problems
Figure 3

large begin{aligned} \Here;the;upward;direction;is;taken;as;the;positive;X-direction&. \The;point;of;projection;of P;is;taken;as;x_{0}=0;at;t=0. \Let;the;two;stones;meet;at;time;t;at;the;point;x_{t}.;Then;the;position;co-&ordinate;x_{t};is;the;same;for;both;P;and;Q. \x_{t}&=x_{0}+v_{0}t+frac{1}{2}at^{2} \Here,;for;the;stone;P,;x_{0}=0,;v_{0}=30;m/s,;&a=-9.8;m/s^{2}. \Similarly,;for;the;stone;Q,;x_{0}=90,;v_{0}=-30;m/s,;&a=-9.8;m/s^{2}. \ie.,;;0+30t-frac{1}{2}times 9.8t^{2}&=90-30t-frac{1}{2}times 9.8t^{2} \therefore 60t&=90 \or,;;t&=frac{90}{60}=underline{underline{1.5;seconds}} \\The;distance;travelled;by;the;first;stone;during;this;interval,;left ( x_{t}-x_{0} right )&=v_{0}t+frac{1}{2}at^{2} \&=30times 2-frac{1}{2}times 9.8times 1.5^{2} \&=60-11.03 \&=underline{underline{48.97;m}} end{aligned}
large begin{aligned} \ie.,;the;two;stones;will;meet;at;a;height;of;48.97;m;from;the;point;of;projection;of;P,;after;1.5;seconds. end{aligned}

Problem 18:

An object moving with an initial velocity of 15 m/s, has a uniform acceleration of 3 m/s2. Find the distance travelled by the object in the 10th second of its motion.

Solution:

large begin{aligned} \Given,;v_{0}=15;m/s,;a=3;m/s^{2},;n=10 \We;know,;;s_{n}&=v_{0}+frac{1}{2}a(2n-1) \therefore ;substituting,;we;get,;s_{10}&=15+frac{1}{2}times 3times (2times 10-1) \&=15+28.5 \&=underline{underline{43.5;m}} end{aligned}

Tutorial 19:

A stone is thrown vertically up with a velocity of 10 m/s. Find (a). The maximum height reached by the stone. (b). the time taken to reach the maximum height. (c). the velocity with which it touches the ground. (d). the time taken to reach the ground.

Answer:

large begin{aligned} \Given,;v_{0}=10;m/s,;a=-g=-9.8;m/s^{2}. \(a).;Let;s=h,;is;the;maximum;height;&reached;by;the;stone,;where;its;velocity;becomes;0. \ie.,;v_{t}=0 \We;have,;v_{t}^{2}&=v_{0}^{2}+2as \therefore s&=frac{v_{t}^{2}-v_{0}^{2}}{2a} \therefore h&=frac{0-10^{2}}{2times -9.8} \&=underline{underline{5.1;m}} \\(b).;Let;t;be;the;time;taken;to;reach;the;&maximum;height. \We;have,;v_{t}&=v_{0}+at \or,;t&=frac{v_{t}-v_{0}}{a} \&=frac{0-10}{-9.8} \&=underline{underline{1.02;s}} \\(c).;Let;v_{t};is;the;velocity;acquired;by;the&;stone;on;reaching;the;ground. \Here,;v_{0}=0,;a=g=9.8;&m/s^{2},;s=h=5.1;m,;v_{t}=? \We;know,;v_{t}^{2}&=v_{0}^{2}+2as \&=0+2times 9.8times 5.1 \&=99.96 \therefore v_{t}&=sqrt{99.96} \&approx underline{underline{10;m/s}},;which;is;equal;to;the;velocity;of;projection. end{aligned}

large begin{aligned} \\(d).;Let;T;be;the;total;time;taken;by;the;stone;to;&reach;the;ground. \For;the;entire;displacement,;s=0,;t=T,;&v_{0}=10;m/s, \;a=-g=-9.8;m/s^{2}&(initial;motion;is;upward) \We;have,;s&=v_{0}t+frac{1}{2}at^{2} \ie.,;;0&=v_{0}T+frac{1}{2}aT^{2} \&=v_{0}+frac{1}{2}aT \therefore T&=frac{-2v_{0}}{a} \ie.,;;T&=frac{-2times 10}{-9.8} \&=underline{underline{2.04;s}} end{aligned}

Question 20:

A stone thrown vertically upward with a speed of 25 m/s from the top of a building returns to the earth in 8 seconds. Find the height of the building.

Solution:

large begin{aligned} \Given,;s=-H,;t=8;s,;&a=-9.8;m/s^{2},;v_{0}=25;m/s \We;know,;;;s&=v_{0}t+frac{1}{2}at^{2} \Substituting,;we;get,;;-H&=v_{0}t-frac{1}{2}at^{2} \therefore H&=-v_{0}t+frac{1}{2}at^{2} \&=-25times 6+frac{1}{2}times 9.8times 8^{2} \&=underline{underline{163.6;m}} \ie.,;the;height;of;the;buil&ding;is;163.6;metres. end{aligned}

Problem 21:

A ball dropped into a well hits the water surface in 6 seconds. How deep is the well and with what velocity the ball hit the water surface?

Answer:

large begin{aligned} \Given,;t=6;s,;v_{0}=0,;a=g=9.8;&m/s^{2},;s=?,;v_{t}=? \We;know,;s&=v_{0}t+frac{1}{2}at^{2} \&=0+frac{1}{2}times 9.8times 6^{2} \&=underline{underline{176.4;m}} \\Again,;v_{t}&=v_{0}+at \&=0+9.8times 6 \&=underline{underline{58.8;m/s}} \ie.,;the;well;is;176.4;m;deep;and;the;&ball;hits;the;water;surface;with;a;velocity;58.8;m/s. end{aligned}

Problem 22:

A ball is dropped into a well of depth 200 m. The sound of splash is heard after 7 seconds. Find the velocity of sound in air.

Solution:

large begin{aligned} \Given,;v_{0}=0,;a=g=9.8;&m/s^{2},;s=200;m \Let;t;be;the;time;taken;for;stone;to;reach;the;&surface;of;water. \We;know,;s&=v_{0}t+frac{1}{2}at^{2} \substituting,;;200&=0+frac{1}{2}times 9.8times t^{2} \therefore t^{2}&=40.82 \therefore t&=sqrt{40.82} \&=6.39;s \therefore;the;time;taken;by;the;sound;to;reach;the;surface;of;well&=7-6.39=0.61;s \therefore Velocity;of;sound&=frac{Distance;travelled}{time} \&=frac{200}{0.61} \&=underline{underline{327.9;m/s}} end{aligned}

Example 23:

A car travelling at 45 km/h is brought to rest with uniform retadation in 15 seconds. Find the retardation.

Answer:

large begin{aligned} \Given,;v_{0}=45;km/h=45times frac{5}{18}=&12.5;m/s,;v_{t}=0,;t=15;s. \We;know,;v_{t}&=v_{0}+at \therefore a&=frac{v_{t}-v_{0}}{t} \&=frac{0-12.5}{15} \&=underline{underline{-0.83;m/s}} \ie.,retardation;is;0.83;m/s. end{aligned}

Problem 24:

A ball is dropped from the top of a building of 180 metres high and at the same time another ball is projected vertically upwards from the ground with a velocity of 45 m/s. Calculate when are where the two balls will meet.

Answer:

motion related problems
Figure 4

large begin{aligned} \For;stone;P, \v_{0}=45;m/s,;s=x,;&a=-9.8;m/s^{2} \We;know,;s&=v_{0}t+frac{1}{2}at^{2} \substituting,;we;get,;x&=45t+frac{1}{2}times -9.8times t^{2} \implies;;x&=45t-4.9 t^{2};;;;;;--------(1) \\For;stone;Q, \v_{0}=0;m/s,;a=9.8;&m/s^{2},;s=180-x \Again;substituting;in;the;equation,;s&=v_{0}t+frac{1}{2}at^{2} \we;get,;180-x&=0times t+frac{1}{2}times 9.8times t^{2} \implies;;180-x&=4.9 t^{2};;;;;;;;;;;;;;--------(2) \\Now,;eqn;(1)+(2);implies ;x+180-x&=45t-4.9 t^{2}+4.9 t^{2} \ie.,;180&=45t \or,;t&=frac{180}{45}=underline{underline{4;s}} \\Substituting;the;value;of;t;in;eqn;(1)implies ;x&=45times 4-4.9times 4^{2} \&=underline{underline{101.6;m}} \\ie.,;the;balls;will;meet;4;seconds;after;start;at;&a;height;of;101.6;metres;above;the;ground. end{aligned}

Question 25:

An object falls freely from rest from the top of a building describes 54 metres in the last second of its fall. Find the height of the building.

Answer:

large begin{aligned} \Given,;s_{n}=54;m,;v_{0}=&0,;a=g=9.8;m/s^{2} \We;know,;s_{n}&=v_{0}+frac{1}{2}a(2n-1) \Subsituting,;we;get,;;54&=0+frac{1}{2}times 9.8times (2n-1) \ie.,;(2n-1)&=frac{54times 2}{9.8} \&=11.02 \therefore n&=frac{11.02+1}{2} \&=6.01;s \\Hence;the;object;takes;6.01;seconds&;to;reach;the;ground. \Now,;the;height;of;the;tower;is,;;s&=v_{0}t+frac{1}{2}at^{2} \&=0+frac{1}{2}times 9.8times 6.01^{2} \&=underline{underline{176.99;m}} \\Thus;the;height;of;the;building;is;&176.99;metres. end{aligned}

Problem 26:

A stone is dropped from a point 5.4 m above a window 1.8 m high. Find the time taken by the stone to pass against the window.

Answer:

kinematic motion problems
Figure 5

large begin{aligned} \Given,;v_{0}=;0,;a=g=&9.8;m/s^{2} \The;time;taken;to;travel;against;the;window,\;t=time;taken;to;travel;PR(t_{1})&-time;taken;to;travel;PQ(t_{2}) \\underline{mathbf{For;PR}} \s=5.4+1.8=7.2;m,;&v_{0}=;0,;a=g=9.8;m/s^{2} \We;have;s&=v_{0}t+frac{1}{2}at^{2} \ie.,7.2&=0+frac{1}{2}times 9.8times t_{1}^{2} \therefore t_{1}^{2}&=1.47 \ie.,;t_{1}&=sqrt{1.47} \&=1.21;s \\underline{mathbf{For;PQ}} \s=5.4,;v_{0}=;0,;a=g&=9.8;m/s^{2} \We;have;s&=v_{0}t+frac{1}{2}at^{2} \ie.,5.4&=0+frac{1}{2}times 9.8times t_{2}^{2} \therefore t_{2}^{2}&=1.10 \ie.,;t_{1}&=sqrt{1.10} \&=1.05;s \\therefore time;taken;by;the;stone;to;travel;across;the;window,t&=t_{1}-t_{2} \&=1.21-1.05 \&=underline{underline{0.16;s}} end{aligned}

Problem 27:

An object describes 6-metre distance in 5th second and 18 metres in the 7th second. If the motion is uniformly accelerated, how far will it travel in the next 4 seconds?

Solution:

large begin{aligned} \Given,;s_{5}=6;m,s_{7}=18;m \we;know,;s_{n}&=v_{0}+frac{1}{2}a(2n-1) \In;the;5^{th};second,;6&=v_{0}+frac{1}{2}a(2times 5-1) \ie.,6&=v_{0}+frac{1}{2}atimes 9;;;;;;------(1) \Similarly,;In;the;7^{th};second,;18&=v_{0}+frac{1}{2}a(2times 7-1) \ie.,18&=v_{0}+frac{1}{2}atimes 13;;;;;;------(2) \eqn(2)-(1)implies12&=0+frac{1}{2}a(13-9) \therefore a&=underline{underline{6;m/s^{2}}} \\Also;substituting;in;eqn;(1),;we;get,;6&=v_{0}+frac{1}{2}times 6times 9 \therefore v_{0}&=underline{underline{-21;m/s}} \\Now,;the;velocity;of;the;object;at;the;end;of;7;sec;is,;v_{t}&=v_{0}+at \&=-21+6times 7 \&=underline{underline{21;m/s}} \\Now;the;distance;travelled;in;next;4;seconds,;s&=v_{0}t+frac{1}{2}at^{2} \Here;v_{0}=21;m/s \therefore s&=21times 4+frac{1}{2}times 6times 4^{2} \&=underline{underline{132;m}} end{aligned}

Problem 28:

A parachutist bails out from a helicopter and after dropping through a distance of 50 metre opens the parachute and decelerates at 4 m/s2. If he reaches the ground with a speed of 3 m/s, how long is he in the air? At what height did he bail out from the helicopter?

Answer:

large begin{aligned} \When;the;parachutist;is;dropped;from;the;&copter;before;opens;the;parachute,;\s=50;m,;v_{0}=0,;a=g=9.8;m/s^{2},;t_{1}=? \We;know,;v_{t}^{2}&=v_{0}^{2}+2as \implies;v_{t}^{2}&=0+2times 9.8times 50 \&=980\therefore v_{t}&=sqrt{980}=31.3;m/s \\Again;v_{t}&=v_{0}+at \implies 31.3&=0+9.8times t_{1} \therefore;t_{1}&=frac{31.3}{9.8}=3.19;s \\When;the;parachute;opens,\;v_{0}=31.3;m/s,;v_{t}=3;m/s,;a=-4;m/&s^{2},;t_{2}=?,;s=? \Now,;v_{t}&=v_{0}+at \implies3&=31.3-4times t_{2} \therefore t_{2}&=7.08;s \Now,;Total;time,;t_{1}+t_{2}&=3.19+7.08\&=underline{underline{10.27;s}} \\Again,;v_{t}^{2}&=v_{0}^{2}+2as \therefore s&=frac{v_{t}^{2}-v_{0}^{2}}{2a} \&=frac{3^{2}-31.3^{2}}{2times -4} \&=121.3;m \therefore Total;height&=50+121.3 \&=underline{underline{171.3;m}} end{aligned}

Problem 29:

An object travels a distance of 4 metres in 4 seconds and 4.4 metres in next 8 seconds. What will be the velocity of the object at the end of 13th second from the start?

Solution:

large begin{aligned} \Given,;s_{1}=4;m,s_{2}=4.4;m,&;t_{1}=4;s,;t_{2}=8;s \Total;distance;covered,;s&=4+4.4=8.4;m \Total;time;taken,;t&=4+8=12;s \We;know,;s_{1}&=v_{0}t_{1}+frac{1}{2}atimes t_{1}^{2} \Substituting,;;4&=v_{0}times 4+frac{1}{2}atimes 4^{2} \Simplifying,;;1&=v_{0}+2a;;;;;;-------(1) \\Again,;;s&=v_{0}t+frac{1}{2}at^{2} \Substituting,;8.4&=v_{0}times 12+frac{1}{2}times atimes 12^{2} \Simplifying,;;8.4&=12left ( v_{0}+6a right ) \implies;0.7&=v_{0}+6a;;;;;;-------(2) \\eqn(1)-(2);implies 0.3&=0-4a \implies a&=-.075;m/s^{2} \substituting;in;eqn(1)implies 1&=v_{0}-2times 0.075 \implies v_{0}&=1.15;m/s \\Now,;the;velocity;of;the;object&;at;the;end;of;13^{th};second;is, \v_{t}&=v_{0}+at \&=1.15-.075times 13 \&=underline{underline{0.175;m/s}} end{aligned}

Hope you understood the workout examples and problems related to the kinematic equations of motion.

Stay tuned with HelpYouBetter to learn more interesting topics and related concepts like the derivation for three equations of motion, equations of vertical motion under gravity, etc.

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