December 2023

17.A car accelerates from rest at a constant rate α for sometime and reaches a maximum velocity. Then at decelerates rate β to come to rest. If the time lapsed is ‘t’ ,then the maximum velocity acquired by the will be

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solution: The car starts from rest (u)=0 acceleration (a)=α to reach maximum velocity “V1“ then ,the car decelerates with (a)=-β final velocity (V2)=0 for the first case ,V1=0+αt1,t1=V1/α for the second case, initial velocity(u)=V1 with deceleration (a)=-β final velocityV2 =0 0=V1-βt2 V1=βt2,t2=V1/β Total time=V1/α+V1/β=V1(β+α)/αβ=V1=(αβ/α+β)t V1=(αβ/α+β)t

15.An object projected vertically upwards with an initial velocity 40m/s from the ground(g=10m/s2).The time taken by the object to reach the maximum height

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solution: The object projected vertically upwards an initial velocity(u)=40m/s if the object to reach the maximum height, the final velocity will be zero. from kinematic equations ,we have the formula, (v)2-(u)2=2as here , v=0,u=40m/s, a=-g=10m/s2 0-40*40=2(-10)S -1600=-20S S=80m here, distance is nothing but maximum height .

13.In a car race, in a very busy road , a skilled driver has managed to drive his car with uniform velocity of 5m/s for 10 minutes, then average speed of the car in first 2 minutes

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solution: in the question ,the driver managed his car with 5m/s. the time is 10minutes the car is moving with uniform velocity with 5m/s that means ,the car has travelled with 5m/s through out the journey. i.e at 2minutes and 10minutes also ,the car has same velocity. the average speed of the car is same.i.e.5m/s …

13.In a car race, in a very busy road , a skilled driver has managed to drive his car with uniform velocity of 5m/s for 10 minutes, then average speed of the car in first 2 minutes Read More »

12.A car moving with a speed of 20m/s can be stopped by applying brakes, after travelling a distance of 2m.The retardation of the car is

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solution: car is moving with a speed(u)=20m/s it can stopped by brakes ,that means final velocity (v)=0 in mean time ,it travelled a distance (S)=2m. by given information ,initially it has initial velocity and then comes rest .so, the car is in retarded motion. we have the formula, V2-u2=2aS 0-(20)2 =2a(2) -400=4a a=-100m/S2 The car …

12.A car moving with a speed of 20m/s can be stopped by applying brakes, after travelling a distance of 2m.The retardation of the car is Read More »

11.A particle initially at rest starts moving with a uniform acceleration “a”. The ratio of the distances covered by it in first and the first 3s is

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solution: a particle starts from rest (u)=0 the distance covered in first second (t)=1sec we have the formula S=ut+1/2at2 here ,u=0,t=1s,a=a S=0+1/2a(1)2 S1=1/2a the distance covered in 3sec S=ut+1/2at2 S=0+1/2a(3)2 S2=9/2a S1 :S2=1:9 the ratio of the distances covered by body is 1:9

10.If a bus accelerates from rest for time t1,at a constant rate α and then retards at a constant rate β for time t2 and comes to rest, then t1/t2=

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solution: a bus initial velocity u=0 for time =t1 acceleration = α after time t1 ,the velocity will be v1=αt1 this velocity will become initial velocity for retarded motion, u=αt1 then retards with acceleration =-β for time =t2 this is given information , for time t2 ,the body retards v2= αt1-βt2 but v2 is nothing …

10.If a bus accelerates from rest for time t1,at a constant rate α and then retards at a constant rate β for time t2 and comes to rest, then t1/t2= Read More »

8.The velocity of a particle moving with a uniform acceleration of 10m/s2 after 5s is 50m/s. Its velocity at t= 4s is

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solution: In the above problem, acceleration of a particle is (a)=10m/s2 time=5s final velocity=50m/s but it is moving with uniform acceleration. v=u+at 50=u+10*5 u=0m/s i.e it starts with zero initial vvelocity then we have to calculate velocity at 4s for this a=4m/s2 t=4s u=0m/s v=0+10*4 v=40m/s the velocity at 4s is 40m/s.

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