You're trying to stop a car that is rolling down-hill. Gravity is pulling it forward with a force of 200 N. If the car has a mas
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Answer:
The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
Explanation:
The centripetal acceleration is given by:
Where:
: is the tangential speed = 9.50 m/s
r: is the distance = 6.00 m
Hence, the centripetal acceleration is:
Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
I hope it helps you!
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time,
(a) Acceleration,
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration,
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.