How to determine the acceleration of an object given the distance that the object has travelled in a time interval?

Question

The motion of a ball rolling from rest down a plank is measured by marking its position each second. From $t=3s$ to $t=4s$, the ball travels $0.56m$. Calculate the magnitude of the acceleration of the ball.
I calculated the average velocity of the ball between $t=3s$ to $t=4s$ like so:
$Δv=frac{Δs}{Δt}$ = $frac{0.56}{4-3}$ = $0.56m/s$
and then I proceed to:
$Δa=frac{Δv}{Δt}$ = $frac{0.56-0}{4}$ = $0.14m/s^2$
However this is incorrect. I think I may have some idea where I went wrong, for example $Δv$ is the average velocity between $t=4$ and $t=3$ but looking at the acceleration equation

$v$ in the above equation is the final instantaneous velocity? Does the final velocity needs to be instantaneous?Also if I were to visualise this how can i draw a velocity time graph?

bemjanim · Answer

You must use the equations of uniform acceleration. Your formula for velocity is wrong. $v=frac{ds}{dt}$ not $frac{Delta s}{Delta t}$. Your formula is for constant velocity, so won't work here. Use instead the formula $s=ut + frac{1}{2} a t^2$, which simplifies to $s=frac{1}{2}at^2$ because $u=0$.

Answered by bemjanim on July 6, 2021

gandalf61 · Answer

The ball travels $0.56$ m in the $1$ second interval between $t=3$ and $t=4$ so its average speed over this interval is $0.56$ m/s.
To find its acceleration, notice that a constant acceleration $a$ means that its speed at $t=3$ is $3a
$ and at $t=4$ it is $4a$. So the ball’s average speed between these two times is $frac 7 2 a$. So we have
$displaystyle frac 7 2 a = 0.56$
$Rightarrow a=0.16$ m/s/s

How to determine the acceleration of an object given the distance that the object has travelled in a time interval?

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