Answer to Question #128853 in Mechanics | Relativity for Ariyo Emmanuel

Question

A particle travels along a straight line with a velocity of v= (20 − 0.05s^2) m/s, where s is in meters. Determine the acceleration of the particle at s = 15 m.

Answer 1

Velocity v=(20-0.05s$^2$ )m/s

at s=15m ,

velocity v=(20-0.05$\times 15^2)$

v=18.75m/s

as we know acceleration is the rate of change of velocity

a=$\frac{dv}{dt}$

=$\frac{d(20-0.05s^2)}{dt}$

= 0-0.05$\times 2s\times \frac{ds}{dt}$

=-0.05$\times 2 \times 15\times v$ ( since velocity is rate of change of distance)

= $-1.5\times18.75$

= - 28.125m/s$^2$

Hence the acceleration at s=15m is - 28.125m/s$^2$