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Question

A toy car accelerates from rest at 2 ms−2 for 3 seconds. Afterwards the car begins to slow down. It travels a combined total of 27 meters before coming to rest. What is its rate of acceleration when stopping and what is the total time of the motion?

A
a=2 ms2, t=6 s
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B
a=1 ms2, t=9 s
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C
a=2 ms2, t=9 s
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D
a=1 ms2, t=6 s
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Solution

The correct option is B a=1 ms2, t=9 s
Let the initial velocity of toy car be 0 ms1, velocity to which the toy car is accelerated be v1 ms1
Given: time of acceleration t1=3 ms1
We can divide this question into two parts. One in which the toy car accelerates from 0 ms1 to v1 ms1
Next, when it decelerates from v1 ms1 to 0 ms1
Let the total distance travelled be 's' and total time taken be 't'.
Case I: when the car accelerates
Initial velocity, u1=0 ms1
acceleration, a1=2 ms2 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 ms1

Case II: When the car is retarding from 6 ms1 to 0 ms1
Final velocity, v2=0 and initial velocity, u1=6 ms1
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 ms2
We know, v2=u2+a2t2
0=6+(1)t2
t2=6 s

Total time:
t=t1+t2
t=3+6=9 s

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