A particle starts from rest and moves with constant acceleration alpha for time t_1, then decelerates uniformly to rest in time t_2. The ratio of distance covered during acceleration to that during deceleration is:

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NEET Physics QUESTION #7306

Question 1

A particle starts from rest and moves with constant acceleration $\alpha$ for time $t_1$, then decelerates uniformly to rest in time $t_2$. The ratio of distance covered during acceleration to that during deceleration is:

$\dfrac{t_1}{t_2}$✔️
$\dfrac{t_2}{t_1}$
$\dfrac{t_1^2}{t_2^2}$
$\dfrac{t_2^2}{t_1^2}$

Correct Answer Explanation

Max velocity reached: $v_{\max} = \alpha t_1$

Distance in acceleration phase: $s_1 = \dfrac{1}{2}\alpha t_1^2$

Distance in deceleration phase: $s_2 = \dfrac{1}{2}v_{\max}\,t_2 = \dfrac{1}{2}\alpha t_1 t_2$

Ratio: $\dfrac{s_1}{s_2} = \dfrac{\frac{1}{2}\alpha t_1^2}{\frac{1}{2}\alpha t_1 t_2} = \mathbf{\dfrac{t_1}{t_2}}$

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