The correct option is B a=−1 ms−2, t=9 s
Let the initial velocity of toy car be 0 ms−1, velocity to which the toy car is accelerated be v1 ms−1
Given: time of acceleration t1=3 ms−1
We can divide this question into two parts. One in which the toy car accelerates from 0 ms−1 to v1 ms−1
Next, when it decelerates from v1 ms−1 to 0 ms−1
Let the total distance travelled be 's' and total time taken be 't'.
Case I: when the car accelerates
Initial velocity, u1=0 ms−1
acceleration, a1=2 ms−2 and final velocity be v1
Using s1=u1t1+12a1t21
s1=0+12×2×32=9 m
Now, using v1=u1+a1t1,
v1=0+2×3=6 ms−1
Case II: When the car is retarding from 6 ms−1 to 0 ms−1
Final velocity, v2=0 and initial velocity, u1=6 ms−1
a2 be the acceleration and s2 be the distance covered.
Using v22=u22+2a2s2
02=62+2a2s2
s2=−18a2
Total distance travelled, s=27 m
s=s1+s2=9+−18a2 = 27
⇒ a2=−1 ms−2
We know, v2=u2+a2t2
∴0=6+(−1)t2
⇒t2=6 s
∴ Total time:
t=t1+t2
t=3+6=9 s