Question

# A body starts from rest with acceleration 2 m/s2 till it attains the maximum velocity, then retards to rest with 3 m/s2. If total time taken is 10 seconds then maximum speed attained is

A
12 m/s
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B
8 m/s
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C
6 m/s
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D
4 m/s
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Solution

## The correct option is A 12 m/sI - Method Let the maximum velocity attained at time (t′) be vmax. By first equation of motion vmax=u+at′ ⇒vmax=2t′ ...(i) Now for the second part of the journey body comes to rest (v=0) and the time will be (10−t′) as the total journey time is 10 s, u=vmax, a=−3 m/s2. By first equation of motion 0=vmax−3(10−t′) ...(ii) putting the value of vmax in equation (ii) we get, 2t′=3(10−t′), ⇒t′=6 s vmax=2t′=2×6=12 m/s II−Method As we know that slope (tanθ) of velocity-time (v−t) graph gives acceleration. tanα=2, tanβ=3 2=vt, 3=v10−t v=2t, v=30−3t 30−3t=2t⇒t=6 sec v=12 m/s

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