Fred goes down a 5.0-m slide at a playground in 2.0 s. He starts from rest and accelerates uniformly. Show that his acceleration while on the slide is 2.5 m/s2.
d = 0.5*a*T^2.
5 = 0.5a*2^2,
a =
To solve this problem, we can use the formula for uniformly accelerating motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 0 m/s (Fred starts from rest)
v = ? (final velocity)
a = ? (acceleration)
t = 2.0 s (time taken to slide down)
We want to find the acceleration, so let's rearrange the formula to solve for a:
a = (v - u) / t
Since Fred starts from rest, his initial velocity u is 0 m/s. Therefore, the formula becomes:
a = (v - 0) / t
a = v / t
Now, we need to find the final velocity v. We can use another formula:
v = u + at
Since Fred starts from rest, his initial velocity u is 0 m/s. The formula simplifies to:
v = at
We can substitute this expression for v in the acceleration formula:
a = (at) / t
Simplifying further:
a = (a * t) / t
The t's cancel out:
a = a
This confirms that the acceleration while on the slide is a.
Hence, we have shown that Fred's acceleration while on the slide is 2.5 m/s^2.
To determine Fred's acceleration while on the slide, we can use the kinematic equation that relates distance (d), initial velocity (vi), acceleration (a), and time (t):
d = vi * t + (1/2) * a * t^2
Given:
- Distance (d) = 5.0 m
- Time (t) = 2.0 s
- Initial velocity (vi) = 0 m/s (Fred starts from rest)
Plugging in these values into the equation, we have:
5.0 m = 0 m/s * 2.0 s + (1/2) * a * (2.0 s)^2
Simplifying the equation, we get:
5.0 m = 0 + a * 2.0 s^2
5.0 m = 2.0 s^2 * a
Now, we can solve for acceleration (a) by rearranging the equation:
a = (5.0 m) / (2.0 s^2)
a = 2.5 m/s^2
Therefore, Fred's acceleration while on the slide is 2.5 m/s^2.