Answer: 3 ft
Explanation:
Given:
- the ball is initially at rest before it start moving at constant acceleration.
a = constant
u = initial speed = 0
Using the formula for displacement.
d = ut + 0.5at^2 .....1
Since u = 0, equation 1 becomes;
d = 0.5at^2 .....3
making a the subject of formula
a = 2d/t^2 ....2
Where; a = acceleration, d = distance travelled, t = time taken, u = initial speed
d = 1 ft, t =1 s
Substituting into equation 2 above.
a = (2×1)/1^2 = 2ft/s^2
Since the acceleration is 2ft/s^2, at t= 2 sec the distance it would have covered will be:
Using equation 3;
d = 0.5(2 × 2^2) = 0.5(8) = 4ft
therefore, the distance travelled between t = 1 to t= 2
Is :
d2 = d - d1
Where d = total distance (t=0 to t= 2) =4 ft
d1 = distance covered between (t=0 to t=1) = 1 ft
d2 = 4 - 1 = 3ft
The distance traveled between time 1s and time 2s is 3 ft.
The distance traveled by the ball at time t is calculated as follows;
[tex]s = ut + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2}[/tex]
when the distance, s = 1 ft and time, t = 1 s
[tex]a = \frac{2(1)}{(1)^2} \\\\a = 2 \ ft/s^2[/tex]
When the time of motion, t = 2s, the distance traveled is calculated as;
[tex]s = \frac{1}{2} (2) (2)^2\\\\s = 4 \ ft[/tex]
The distance traveled between t =1 and t = 2 is calculated as follows;
[tex]s = s_2 - s_1\\\\s = 4 \ ft - 1\ ft\\\\s = 3 \ ft[/tex]
Learn more about change in distance here: https://brainly.com/question/2854969
