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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 metres before coming to rest. What is its rate of retardation and what is the total time of the motion?

A
a=1 ms2, t=9 s
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B
a=2 ms2, t=6 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 A a=1 ms2, t=9 s
Given:
Time of acceleration t1=3 ms1,
initial velocity of toy car be 0 ms1

Let,
velocity to which the toy car is accelerated be v1 ms1,
Total distance travelled be 's',
Total time taken be 't'.

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

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 retardation 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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