NSEP 2015 Problem 8 | One Dimensional Motion

A car moving along a straight road at a speed of u m/s applies brakes at $t=0$ seconds. The ratio of distances travelled by the car during $3^{rd}$ and $8^{th}$ seconds is $15:13$. The car covers a distance of $0.25m$ in the last second of it's travel. Therefore, the acceleration a (in $m/s^2$) and the speeed u($m/s$) of the car are respectively,

a) $-0.1,16$

b) $-0.2,12$

c) $-0.5,20$

d) $-0.1,16$

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(c) $-0.5,20$

We know that displacement in $t$ sec is given by,

$$ s = ut +\frac{1}{2} a t^2$$

where $u$ is the initial velocity and $a$ is the acceleration. This will be negative for deceleration.

Using this idea we can just find the expression for the displacement in the $n^{th}$ second, for which the information is given to us.

Also, the car come to rest and at the second second it travels a distance of $0.25m$. WE can think it in opposite manner that with an acceleration of $a$ and initial velocity $0m/s$, the car moves a distance of $0.25m$ , hence,

$$0.5 =\frac{1}{2} a (1^2)$$

So, the displacement in $n^{th}$ second is,

$$ s_1 = ut+\frac{1}{2} a n^2 $$

and for the $(n-1)^{th}$ second the displacement is,

$$ s_2= ut+\frac{1}{2} a (n-1)^2 $$

Hence, we have

$$ s_n = s_1-s_2 = u + a(n-\frac{1}{2}) $$

we have,

$$ \frac{15}{13} = \frac{s_3}{s_8} = \frac{u-\frac{5a}{2}}{u - \frac{15a}{2}} $$

Also we have $0.5 =\frac{1}{2} a (1^2)$

Solving this two gives us $a = 0.25m/s^2 $. Putting this in first equation gives us $u = 20 m/s $.

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