A golfer hits a golf ball at an angle of 30 degrees from the ground, with an initial velocity of 120 ft/sec. It lands on the gro
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Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Initial velocity (Vi) = 25 m/s
acceleration (a) =
time interval (t) = 5 sec
let us assume that final velocity after 5 sec be Vf
As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e.
Hence,
so, the velocity of bicyclist will be 5 m/s after 5 sec