Question

A rolling ball has an initial velocity of 1.6 m/s.a. If the ball has a constant acceleration of 0.33 m/s2 in the same direction as its movement,what is its speed after 3.6 s?b. How far did the ball travel?

Answer 1

Since we are dealing with constant acceleration, we can apply the equations of motion.

Given

Vo = Initial velocity

a = accelaration

t = time

Vo = 1.6 m/s

a = 0.33 m/s2

t = 3.6 s

Procedure

`\begin{gathered} V_f_{}=v_o+a\cdot t \\ V_f=1.6m/s+0.33m/s^2\cdot3.6 \\ V_f=2.79\text{ m/s} \end{gathered}`

Now, for the distance:

`\begin{gathered} x=v_ot+(1)/(2)at^2_{} \\ x=1.6\text{ m/s}\cdot3.6\text{ s+}(1)/(2)\cdot0.33\cdot3.6^2 \\ x=7.9\text{ m} \end{gathered}`