A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If acceleration is constant, what distance does it travel?
A. 1.25 m
B. 2.5 m
C. 5 m
D. 10 m
Answer is B. 2.5 m
I can solve this for displacement real easy using the following equation for uniform accelerated motion x=Vi(t)+1/2(a)(t^2) (answer would be zero), but I have no clue how to approach solving this for distance.
The best I could come up with:
Distance = Average Speed * time
initial speed is = 5 m/s
final speed is = 5 m/s
Time = 1/2 second? why???
Average speed = (initial speed+final speed)/2
10/2 = 5 m/s
Distance = 5 m/s * 1/2 second = 2.5 meters
^^^^^^^^^^^^^^^^^^
WHY ARE THEY TAKING TIME as = 1/2 SECOND, so confused as to why time would be 1/2 second
A. 1.25 m
B. 2.5 m
C. 5 m
D. 10 m
Answer is B. 2.5 m
I can solve this for displacement real easy using the following equation for uniform accelerated motion x=Vi(t)+1/2(a)(t^2) (answer would be zero), but I have no clue how to approach solving this for distance.
The best I could come up with:
Distance = Average Speed * time
initial speed is = 5 m/s
final speed is = 5 m/s
Time = 1/2 second? why???
Average speed = (initial speed+final speed)/2
10/2 = 5 m/s
Distance = 5 m/s * 1/2 second = 2.5 meters
^^^^^^^^^^^^^^^^^^
WHY ARE THEY TAKING TIME as = 1/2 SECOND, so confused as to why time would be 1/2 second
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