The displacement and the increase in the velocity of a moving particle in the time interval of t to ( t + 1 ) s are 125 m and 50 m s - 1 , respectively. The distance travelled by the particle in ( t +

Question

The displacement and the increase in the velocity of a moving particle in the time interval of t to ( t + 1 ) s are 125 m and 50 m s - 1 , respectively. The distance travelled by the particle in ( t + 2 ) th s is ___________ m .

TestHub · Accepted Answer

Considering acceleration is constant, we can write

$v = u + a t$

As increase in velocity in $1 s$ is given as $50 m s^{- 1}$ , therefore

$\Rightarrow a = 50 m s^{- 2}$

Now, displacement covered between given time interval is,

$125 = u t + \frac{1}{2} a t^{2}$

$\Rightarrow 125 = u + \frac{a}{2}$

$\Rightarrow u = 100 m s^{- 1}$

Velocity after $1 s$ will be, $v = 100 + 50 = 150 m s^{- 1}$

Therefore, displacement covered in $(t + 1) s to (t + 2) s$ will

$∴ S_{n^{t h}} = v t^{'} + \frac{1}{2} a {(t^{'})}^{2}, where t^{'} = 1 s$

$= 175 m$

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