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Question

# A particle moves along a straight line and its velocity depends on time as v=4t−t2. 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 ms−1 D Acceleration is 4 ms−2at t=0Average velocity →v=∫50vdt∫50dt=∫50(4t−t2)dt∫50dt =[2t2−t33]505=50−12535=253×5=53 ms−1 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∣∣∣+∣∣∣∫54vdt∣∣∣∫50dt =∣∣∣∫40(4t−t2)dt∣∣∣+∣∣∣∫54vdt∣∣∣5 =[2t2−t33]40+[−2t2+t33]545 =(32−643)+[−2(25−16)+13(125−64)]5=135 ms−1 For acceleration: a=dvdt=ddt(4t−t2)=4−2t At t=0,a=4 ms−2 ∴ Options c and d are correct

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