A body starting from rest moves with constant acceleration. ratio of distance covered by body during

English

Gujarati

Hindi

2.Motion in Straight Line

medium

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ sec to that covered in $5\, sec$ is

A

$9/25$

B

$3/5$

C

$25/9$

D

$1/25$

Solution

(a) Distance covered in $5^{th}$ second,

${S_{{5^{th}}}} = u + \frac{a}{2}(2n – 1) = 0 + \frac{a}{2}(2 \times 5 – 1) = \frac{{9a}}{2}$

and distance covered in $5$ second,

${S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times a \times 25 = \frac{{25a}}{2}$

$\therefore \frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}$

Standard 11

Physics

Share

Similar Questions

A car starts from rest and moves with uniform acceleration a on a straight road from time $t = 0$ to $t = T$. After that, $a$ constant deceleration brings it to rest. In this process the average speed of the car is

hard

A particle starts moving from rest in a straight line with constant acceleration. After time $t_0$, acceleration changes its sign (just opposite to the initial direction), remaining the same in magnitude. Determine the time from the beginning of motion in which the particle returns to the initial position.

hard

A body starts with an initial velocity of $10\,ms ^{-1}$ and is moving along a straight line with constant acceleration. When the velocity of the particle is $50\,ms ^{-1}$, the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is $…………\,m/s^{-1}$

hard

The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s $ while travelling through a wooden block of thickness $10\,cm$. Assuming it to be uniform, the retardation will be

easy

(AIIMS-2001)

An automobile travelling with a speed of $60\,\,km/h,$ can brake to stop within a distance of $20 \,m$. If the car is going twice as fast, i.e. $120\, km/h$, the stopping distance will be ……….. $m$

medium

(AIEEE-2004)