Asked by Princess
A 2400-pound car exerts its maximum force of 750 pounds. What will be its minimum time needed to slow it down from 50 m/s to 10 m/s? Determine how far will the car travel in that time.
Answers
Answered by
drwls
deceleration rate =
a = (velocity change)/time
= -F/m = -750 lb/74.5 slugs
= -10.06 ft/s^2
Time = (velocity change)/a
Velocity change must be in ft/s units if you are going to use pounds for mass and slugs for force.
delta V = -40 m/s = -131.2 ft/s
Time = (-131.2 ft/s)/(-10.06 ft/s^2)
= 13.0 sec
Multiply that by the average velocity (30 m/s) for the distance (in meters)
a = (velocity change)/time
= -F/m = -750 lb/74.5 slugs
= -10.06 ft/s^2
Time = (velocity change)/a
Velocity change must be in ft/s units if you are going to use pounds for mass and slugs for force.
delta V = -40 m/s = -131.2 ft/s
Time = (-131.2 ft/s)/(-10.06 ft/s^2)
= 13.0 sec
Multiply that by the average velocity (30 m/s) for the distance (in meters)
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