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Question

A particle moves along a straight line and its velocity depends on time as v=4tt2. Then for first 5 s

A
Average velocity is 253 m
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B
Average speed is 10 ms1
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C
Average velocity is 53 ms1
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D
Acceleration is 4 ms2at t=0
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Solution

The correct options are
C Average velocity is 53 ms1
D Acceleration is 4 ms2at t=0
Average velocity v=50vdt50dt=50(4tt2)dt50dt
=[2t2t33]505=5012535=253×5=53 ms1

For average speed,
let us put v=0, which gives t=0 and t=4 ie velocity changes sign at t=4.
Average speed = 40vdt+54vdt50dt
=40(4tt2)dt+54vdt5
=[2t2t33]40+[2t2+t33]545
=(32643)+[2(2516)+13(12564)]5=135 ms1

For acceleration:
a=dvdt=ddt(4tt2)=42t
At t=0,a=4 ms2
Options c and d are correct

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