Question

If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from 19 m/s to a value approaching zero at s = 39 m, determine the acceleration a of the particle when s = 8 m and show that the particle never reaches the 39-m displacement.

Answer 1

Answer:

Explanation:The rate of decrease of velocity is given with respect to displacement. As v approaches zero when s approaches 39, it can be written as

$\frac{dv}{ds} =-19/39\\m=\frac{dv}{ds}\\Hence\\v=ms+c$

where c=19

$v=-\frac{19}{39} *s+19$

when s=8 we have

$v=-\frac{19}{39} * 8 +19\\v=15.1 m/s$

hence a can be found by the chain rule

$a=\frac{ds}{dt} \frac{dv}{ds}\\a=v*\frac{dv}{ds}\\a=15.1 *-\frac{19}{39}\\a=7.357 m/s^2$

From the equation

$v=-\frac{19}{39} *s+19$

the s in term of t can be found

$v=\frac{ds}{dt} \\\int\limits {v} \, dt =\int\limits {} \, ds\\\int\limits {-\frac{19}{39} *s+19} \, dt=\int\limits {} \, ds\\\\\int\limits^t_0 {} \, dt=\int\limits^s_0 {\frac{1}{-\frac{19}{39} *s+19} } \, ds$

$t=-\frac{39}{19} ln(\frac{s-39}{39} )$

Hence it can be seen that if s approaches 39 the t function becomes undefined hence s is never 39m