Asked by Alif
A car takes up from west and accelerate uniformly at the rate of 4m/s square until it reaches a velocity of 10m/s. It then maintains this speed for 5secs and later decelerate at the rate of 2m/s square till it reaches its destination
a) draw the velocity time graph of the motion
b) compute the time taken for the car to reach the velocity of 10m/s and the time taken decelerate
c) find the total distance covered in the course of this motion.
a) draw the velocity time graph of the motion
b) compute the time taken for the car to reach the velocity of 10m/s and the time taken decelerate
c) find the total distance covered in the course of this motion.
Answers
Answered by
Steve
time to accelerate
10m/s / 4m/s^2 = 2.5s
deceleration takes twice as long, since the rate is half as great.
Now figure the distance using the good old formula
s = v<sub>o</sub>t + 1/2 at^2
for each part of the trip.
10m/s / 4m/s^2 = 2.5s
deceleration takes twice as long, since the rate is half as great.
Now figure the distance using the good old formula
s = v<sub>o</sub>t + 1/2 at^2
for each part of the trip.
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