First slide

Instantaneous acceleration

Question

If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from 20 m/s to a value approaching zero at s = 30 m figure, then acceleration of the particle at s = 15 m is

Moderate

A

2/3 m/s²
B

-2 /3 m/s²
C

20/3 m/s²
D

-20/3 m/s²

Solution

The slope of the line = - 2/3
The equation of line is given by
$\begin{array}{l} (v â 20) = â \frac{2}{3} (s â 0) \\ â´ v = 20 â (2 / 3) s \end{array}$
The velocity at s = 15 m is given by
$V = {(\frac{ds}{dt})}_{s = 15 m} = 20 â (\frac{2}{3}) Ã 15 = 10 m / s$
The acceleration can be obtained by differentiating eq. (1) with respect to time, i.e.,
$\begin{array}{l} a = \frac{dv}{dt} = 0 â \frac{2}{3} \frac{ds}{dt} \\ â´ {(\frac{dv}{dt})}_{s = 15 m} = â \frac{2}{3} {(\frac{ds}{dt})}_{s = 15 m} \\ = â \frac{2}{3} Ã 10 = â \frac{20}{3} m / s^{2} \end{array}$

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