A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

zero

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10 m/s

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50 m/s

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70 m/s

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Solution

The correct option is D 70 m/s
$s_{1}$ be the distance travelled.
$u = 10 m / s, v = 50 m / s$
$v^{2} = u^{2} - 2 a s_{1}$
$2500 = 100 - 2 a s_{1}$
$s_{1} = \frac{2400}{2 a}$
$s_{2}$ be the distance travelled.
When $u = 50 m / s, v = 0 m / s$
$v^{2} = u^{2} - 2 a s_{1}$
$0 = 2500 - 2 a s_{1}$
$s_{1} = \frac{2500}{2 a}$
total distance travelled $= s_{1} + s_{2}$
$= \frac{2500}{2 a} + \frac{2400}{2 a} = \frac{4900}{2 a}$
At initial point $u = 0 m / s$
$v^{2} = 2 a s$
$= 2 a \times \frac{4900}{2 a} = 70 m / s$

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A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?

The correct option is D 70 m/ss1 be the distance travelled.u=10m/s,v=50m/sv2=u2−2as12500=100−2as1s1=24002as2 be the distance travelled.When u=50m/s,v=0m/sv2=u2−2as10=2500−2as1s1=25002atotal distance travelled =s1+s2=25002a+24002a=49002aAt initial point u=0m/sv2=2as=2a×49002a=70m/s