The car starts from rest therefore its initial velocity is zero, i.e., v_i=0vi=0 in case when its acceleration is a_1=2 m/s^2a1=2ms2.
Let the car come to a final velocity v_f=vvf=v. in time t_1t1
Then we can write:
v_f=v_i+a_1t_1vf=vi+a1t1
implies v=0+2t_1⇒v=0+2t1
implies v=2t_1⇒v=2t1
implies t_1=v/2.................(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 v_i=v, v_f=0 and a_2=- 4 m/s^2 (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 t_2.
Thus, we can write:
v_f=v_i+a_2t_2
implies 0=v-4t_2
implies v=4t_2
implies t_2=v/4...............(ii)
Adding equations (i) and (ii), we get.
t_1+t_2=v/2+v/4
t_1 +t_2 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 3 seconds.
implies 3=v/2+v/4
implies 12=2v+v
implies 3v=12
implies v=4 m/s
Hence, the maximum velocity the car attained is 4m/s.
Both the given options in the question are wrong.
