A ball initially moving at 0.5m/s rolls down a straight slope, accelerating at 0.25m/s^2 for 4 sec, how far does the ball travel in this time?

The speed at the end of the 4 seconds is

0.5 + 1.0 = 1.5 m/s

The average speed during the 4 second interval is (0.5 + 1.5)/2 = 1.0 m/s

Multiply that average by 4 s for the distance travelled.

To determine the distance traveled by the ball, we can use the equation:

distance = initial velocity * time + (1/2) * acceleration * time^2

Given:
Initial velocity (u) = 0.5 m/s
Acceleration (a) = 0.25 m/s^2
Time (t) = 4 s

Plugging in the values into the equation:

distance = (0.5 m/s) * (4 s) + (1/2) * (0.25 m/s^2) * (4 s)^2
distance = 2 m + (1/2) * 0.25 m/s^2 * 16 s^2
distance = 2 m + 0.125 m/s^2 * 16 s^2
distance = 2 m + 2 m
distance = 4 m

Therefore, the ball travels a distance of 4 meters in 4 seconds.

To find the distance traveled by the ball, we can use the kinematic equation:

\[d = v_0t + \frac{1}{2}at^2\]

Where:
d = distance traveled
\(v_0\) = initial velocity
t = time taken
a = acceleration

Given values:
\(v_0 = 0.5 \, \text{m/s}\) (initial velocity)
a = 0.25 \, \text{m/s}^2 \) (acceleration)
t = 4 \, \text{s} \) (time taken)

Substituting the values into the equation above:

\[d = (0.5 \, \text{m/s})(4 \, \text{s}) + \frac{1}{2}(0.25 \, \text{m/s}^2)(4 \, \text{s})^2 \]

Simplifying the equation:

\[d = 2 \, \text{m} + \frac{1}{2} \times 0.25 \, \text{m/s}^2 \times 16 \, \text{s}^2\]
\[d = 2 \, \text{m} + (0.5)(4) \, \text{m}\]
\[d = 2 \, \text{m} + 2 \, \text{m}\]
\[d = 4 \, \text{m}\]

Therefore, the ball will travel a distance of \(4 \, \text{m}\) in the given time of \(4 \, \text{s}\).