Question

1. A ball rolls with a constant acceleration of 2.5 m/s^2 for 5.0 s with an initial velocity of 5.0 m/s. Calculate the ball’s distance from the starting point at 1.0 s intervals. Make a position-time graph for the object’s motion. In your response, show what you are given, the equation that you used, any algebra/work required, a table of data, and your graph.

Answer 1

Answer:

Explanation:

Given that the initial velocity of the ball is $5.0 m/s$, the constant acceleration is $2.5 m/s^2$, and the time of motion is $5.0 s$.

Denote the initial velocity be u, constant acceleration by a, and the time by $t. So u= 5.0 m/s, a= 2.5 m/s^2$ and $t= 5.0 s$.

From the equation of motion, the distance, s, covered by the object in time, t, starting with velocity u, having constant acceleration, a, is

$s=ut+\frac 12 at^2$

$\Rightarrow s= 5t+\frac 5 2 t^2\;\cdots(i)$

As the time interval is $1.0 s$

So the time after the 1st interval is 1 s and from equation (i) the distance covered  from the starting point is,

$s_{t=1}=5\times 1 +\frac 5 2 \times1^2=7.5 m$.

The time after the 2nd interval is 2 s and from equation (i) the distance covered from the starting point is,

$s_{t=2}=5\times 2 +\frac 5 2 \times2^2=20 m$.

Similarly, after the 3rd interval, t=3s

$s_{t=3}=5\times 3 +\frac 5 2 \times3^2=37.5 m$.

After the 4th interval, t=4s

$s_{t=3}=5\times 4 +\frac 5 2 \times4^2=60 m$.

And finally, after the 5th interval, t=5s

$s_{t=3}=5\times 5 +\frac 5 2 \times5^2=87.5 m$.

The tabulated data and the position-time graph has been shown in the figure.