A body starts from rest with acceleration 2m/s2 till it attains the maximum velocity, then retards to rest with 3m/s2. If total time taken is 10 seconds then maximum speed attained is
A
12m/s
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B
8m/s
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C
6m/s
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D
4m/s
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Solution
The correct option is A12m/s I - 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 10s, u=vmax, a=−3m/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′=6s vmax=2t′=2×6=12m/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=6sec v=12m/s