Question

a car starts from the rest attain 2.5 m/s square from acceleration . what is the velocity of a car to cover 25 m.? find its time also.

Answer 1

The car starts from rest therefore its initial velocity is zero, i.e., vi=0 in case when its acceleration is a1=2ms2.

Let the car come to a final velocity vf=v. in time t1

Then we can write:

vf=vi+a1t1

⇒v=0+2t1

⇒v=2t1

⇒t1=v2.................(i)

Now when it is again coming to rest its initial velocity is that which it attained when it started from rest i.e., v

Hence, when it is again coming to rest at that period vi=v, vf=0 and a2=−4ms2 (NOTE: The negative sign for acceleration is taken because it is retardation). Let the time which it took for coming to rest from the velocity v be t2.

Thus, we can write:

vf=vi+a2t2

⇒0=v−4t2

⇒v=4t2

⇒t2=v4...............(ii)

Adding equations (i) and (ii), we get.

t1+t2=v2+v4

t1+t2 represents the total time for this trip i.e., starting from rest and then again coming to rest.
And it is given that the total time of trip is 3seconds.

⇒3=v2+v4

⇒12=2v+v

⇒3v=12

⇒v=4ms

Hence, the maximum velocity the car attained is 4ms.

Answer 2

the ans is 4m/s to your answer