Question

a car starting from rest accelerate at the rate of 0.5 m/s² for 20 seconds. what is the velocity achieved by it​

Answer 1

Answer:

If t1 be the time for which the car accelerates at the rate f from rest, the distance traveled in time t1 is 

s=s1=0(t1)+21ft12=21ft12 ....(1) (using formula 

s=ut+21at2)

and the velocity at time t1 is v1=ft1

So, the distance traveled in time t will be s2=v1t=ft1t ...(2)

If t2 be the time for which the car decelerates at the rate f/2 to come rest, the distance traveled in time t2 is given by,

02−v12=2(−f/2)s3   (using formula v2−u2=2as) 

or s3=(ft1)2/f=ft12 ...(3)

using (1),  s3=2s1=2s

Given, s1+s2+s3=15s

or s+ft1t+2s=15s

or ftt1=12s

or 12s=ftt1 ..(4)

(4)/(1)⇒s12s=(1/2)ft12ftt1 or t1=t/6

From (1), s=(1/2)f(t/6)2=ft2/72