A child slides down a hill on a toboggan with an acceleration if 1.5 m/s2^2. If she starts at rest, how far has she traveled in (a) 1.0s,(b) 2.0s, and (c) 3.0s?
You can use the equation (d=Vit + 1/2 a t^2)
So for the first one it would be: d= (0)(1) + 1/2 (1.5) (1^2)
And the answer would be .75m
Each time just plug in the time to 't' and keep all the other information the same.
To find how far the child has traveled, we can use the equation of motion:
distance = initial velocity * time + (0.5 * acceleration * time^2)
Where:
- initial velocity is the velocity at the start of the motion (which is 0 m/s in this case because the child starts at rest)
- time is the duration of the motion
- acceleration is the rate at which the velocity changes
Let's calculate the distance traveled in each case:
(a) When t = 1.0s:
The initial velocity (u) is 0 m/s.
Plugging these values into the equation of motion, we get:
distance = 0 * 1.0 + (0.5 * 1.5 * 1.0^2)
distance = 0 + (0.5 * 1.5 * 1.0)
distance = 0 + (0.75)
distance = 0.75 meters
Therefore, the child has traveled a distance of 0.75 meters in 1.0 second.
(b) When t = 2.0s:
Using the same formula, we have:
distance = 0 * 2.0 + (0.5 * 1.5 * 2.0^2)
distance = 0 + (0.5 * 1.5 * 4.0)
distance = 0 + (0.5 * 6.0)
distance = 0 + 3.0
distance = 3.0 meters
Therefore, the child has traveled a distance of 3.0 meters in 2.0 seconds.
(c) When t = 3.0s:
Using the equation of motion again:
distance = 0 * 3.0 + (0.5 * 1.5 * 3.0^2)
distance = 0 + (0.5 * 1.5 * 9.0)
distance = 0 + (0.5 * 13.5)
distance = 0 + 6.75
distance = 6.75 meters
Finally, the child has traveled a distance of 6.75 meters in 3.0 seconds.
To find the distance traveled by the child in each given time interval, we can use the kinematic equation:
distance = 0.5 * acceleration * time^2
Given:
acceleration (a) = 1.5 m/s^2
time (t) = [1.0s, 2.0s, 3.0s]
(a) For 1.0s:
distance = 0.5 * 1.5 * (1.0)^2
= 0.5 * 1.5 * 1.0
= 0.75 meters
Therefore, the child has traveled 0.75 meters in 1.0 second.
(b) For 2.0s:
distance = 0.5 * 1.5 * (2.0)^2
= 0.5 * 1.5 * 4.0
= 3 meters
Therefore, the child has traveled 3 meters in 2.0 seconds.
(c) For 3.0s:
distance = 0.5 * 1.5 * (3.0)^2
= 0.5 * 1.5 * 9
= 6.75 meters
Therefore, the child has traveled 6.75 meters in 3.0 seconds.