Asked by aayushi
what is the acceleration in m/s2of a body moving at 63 km/h and comes to rest in 40s
Answers
Answered by
Jai
Acceleration is change in velocity over time:
a = (v2 - v1)/t
Initial velocity (v1) is given as 63 km/h. Final velocity (v2) is zero, since it came to rest in 40 s (time).
Note that we should convert km/h to m/s too. Thus,
a = [ 0 - (63 km/h * (1 h / 3600 s) * (1000 m / 1 km)) ] / 40 s
a = ?
Acceleration you'll get is negative, units is m/s^2.
a = (v2 - v1)/t
Initial velocity (v1) is given as 63 km/h. Final velocity (v2) is zero, since it came to rest in 40 s (time).
Note that we should convert km/h to m/s too. Thus,
a = [ 0 - (63 km/h * (1 h / 3600 s) * (1000 m / 1 km)) ] / 40 s
a = ?
Acceleration you'll get is negative, units is m/s^2.
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