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Standard XIII

Physics

2nd Equation of Motion

A particle st...

Question

A particle starts with an initial velocity $2.5 m/s$ along the positive $x -$ direction and it accelerates uniformly at the rate $0.50 {m/s}^{2}$ . The distance travelled by the particle in first two seconds will be

4 m

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5 m

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1 m

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6 m

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Solution

The correct option is D $6 m$
We have, $s = u t + \frac{1}{2} a t^{2} = (2.5) (2) + \frac{1}{2} (0.50) (2)^{2}$
$s = 6 m$
Hence, distance travelled by the particle is $s = 6 m$

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A particle starts with an initial velocity 2.5 m/s along the positive x− direction and it accelerates uniformly at the rate 0.50 m/s2. The distance travelled by the particle in first two seconds will be

The correct option is D 6 mWe have, s=ut+12at2=(2.5)(2)+12(0.50)(2)2 s=6 m Hence, distance travelled by the particle is s=6 m

A particle starts with an initial velocity $2.5 m/s$ along the positive $x -$ direction and it accelerates uniformly at the rate $0.50 {m/s}^{2}$ . The distance travelled by the particle in first two seconds will be

The correct option is D $6 m$
We have, $s = u t + \frac{1}{2} a t^{2} = (2.5) (2) + \frac{1}{2} (0.50) (2)^{2}$
$s = 6 m$
Hence, distance travelled by the particle is $s = 6 m$