Answer:
(a) a = - 201.8 m/s²
(b) s = 197.77 m
Explanation:
(a)
The acceleration can be found by using 1st equation of motion:
Vf = Vi + at
a = (Vf - Vi)/t
where,
a = acceleration = ?
Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)
Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s
t = time = 1.4 s
Therefore,
a = (0 m/s - 282.53 m/s)/1.4 s
<u>a = - 201.8 m/s²</u>
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(b)
For the distance traveled, we can use 2nd equation of motion:
s = Vi t + (0.5)at²
where,
s = distance traveled = ?
Therefore,
s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²
s = 395.54 m - 197.77 m
<u>s = 197.77 m</u>
Answer:
5 N
Explanation:
The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N
Answer:
t = 1.42 s and d = 35.5 m
Explanation:
Given that,
Velocity of a roadrunner is 25 m/s
A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.
We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m
Explanation:
Particle moving in a circular path with a constant speed.
