Question

An object moves with constant acceleration 3.95 m/s2 and over a time interval reaches a final velocity of 11.4 m/s.(a) If its original velocity is 5.70 m/s, what is its displacement during the time interval?m(b) What is the distance it travels during this interval?m(c) If its initial velocity is -5.70 m/s, what is its displacement during this interval?m(d) What is the total distance it travels during the interval in part (c)?m

Answer 1

We can solve this by using the equations of kinematics below

`\begin{gathered} v=u+at \\ \Delta x=ut+(1)/(2)at^2 \\ v^2=u^2+2a\Delta x \end{gathered}`

(a).

`\begin{gathered} a=3.95m/s^2 \\ v=11.4\text{ m/s} \\ u=5.70\text{ m/s} \\ \Delta x=\text{?} \\ v^2=u^2+2a\Delta x \\ 11.4^2=5.70^2+2*3.95*\Delta x \\ 129.96-32.49=7.9\Delta x \\ \Delta x=(97.47)/(7.9) \\ \Delta x=12.3379746835\approx12.34m \end{gathered}`

(b)

Since the object moves in a straight part the distance will also be 12.34 m.

(c)

`\begin{gathered} u=-5.70 \\ v=11.4\text{ m/s} \\ a=3.95 \\ v^2=u^2+2a\Delta x \\ 11.4^2=-5.70^2+2*3.95*\Delta x \\ \Delta x=12.34\text{ m} \end{gathered}`

(d)

The total distance is just the total distance covered without direction. In part c the initial velocity was negative, this means it traveled in the opposite direction therefore that part distance can be calculated below

`\begin{gathered} u=-5.70\text{ m/s} \\ v=0\text{ m/s} \\ a=3.95m/s^2 \\ v^2=u^2+2a\Delta x \\ (0)^2=-5.70^2+2*3.95*\Delta x \\ -32.49=7.9\Delta x \\ \Delta x=-(32.49)/(7.9) \\ \Delta x=-4.11265822785\approx-4.11\text{ m} \end{gathered}`

Then the path where the initial velocity was zero and the final velocity was 11.4 m/s can be calculated as follows

`\begin{gathered} u=0\text{ m/s} \\ v=11.4\text{ m/s} \\ a=3.95m/s^2 \\ v^2=u^2+2a\Delta x \\ 11.4^2=0^2+2*3.95*\Delta x \\ 129.96=7.9\Delta x \\ \Delta x=(129.96)/(7.9) \\ \Delta x=16.4506329114\approx16.45\text{ m} \end{gathered}`

Therefore,

total distance = 4.11 m + 16.45 m = 20.56 m