Question

If the velocity v of a particle moving along a straight line decreases linearly with its displacement from $20m{s^{ - 1}}$ to a value approaching zero at $S = 30m$, determine the acceleration of the particle when $S = 15m$

Answer 1

Hint:In order to solve this question, we will first find the constant ratio of change of velocity with displacement with given data and then again we will use general slope formula for a straight line to find the acceleration in terms of derivative form of acceleration.


Formula used:
Slope for a straight line with given two points say $({x_i},{y_i})\& ({x_f},{y_f})$ where subscript i and f represent initial and final value of given quantity is $slope = \dfrac{{{y_f} - {y_i}}}{{{x_f} - {x_i}}}$


Complete answer:
According to the question we have given that the velocity when $S = 0$ is $u = 20m{s^{ - 1}}$ and decreases linearly to a value of zero $v = 0$ at $S' = 30m$ so finding the slope of this straight line with pairs of points $(0,20)\& (30,0)$ so using the formula $slope = \dfrac{{{y_f} - {y_i}}}{{{x_f} - {x_i}}}$ we get,
$
  slope = \dfrac{{dv}}{{ds}} = \dfrac{{0 - 20}}{{30 - 0}} \\
  \dfrac{{dv}}{{ds}} = - \dfrac{2}{3} \\
 $
Now, as slope of a straight line remains constant and can be calculated using any two given points on the line so, now let one of the point is previous one which is $(30,0)$ and let v be the velocity at $S = 15m$ so this point will be $(15,v)$ so using these two points let’s find the slope and equate it with $slope = - \dfrac{2}{3}$
we get,
$
   - \dfrac{2}{3} = \dfrac{{v - 0}}{{15 - 30}} \\
   \Rightarrow v = 10m{s^{ - 1}} \\
 $
Now, we need to find the acceleration at $S = 15m$ but we know acceleration can be written in derivative form as
$
  a = \dfrac{{dv}}{{dt}} \\
   \Rightarrow a = \dfrac{{dv}}{{dt}}.\dfrac{{ds}}{{ds}} \\
   \Rightarrow a = v\dfrac{{dv}}{{ds}} \\
 $
and at $S = 15m$ we have, $v = 10m{s^{ - 1}}$ and $\dfrac{{dv}}{{ds}} = - \dfrac{2}{3}$ which is constant for a given line.
on putting the values, we get
$
  a = v\dfrac{{dv}}{{ds}} \\
   \Rightarrow a = 10( - \dfrac{2}{3}) \\
   \Rightarrow a = - \dfrac{{20}}{3}m{s^{ - 2}} \\
 $
Hence, the value of acceleration at $S = 15m$ is $a = \dfrac{{ - 20}}{3}m{s^{ - 2}}$



Note:Don’t confuse with choosing velocity and displacement on Y-axis and X-axis, we can choose any quantity on Y and X axis but keeping its same for any calculation further and also remember the basic derivative forms of acceleration and all the basic integration formulas used.