a car starts from reset and accelertates to a speed of 60 mph in 20 secounds. th

Question

a car starts from reset and accelertates to a speed of 60 mph in 20 secounds. the acceleration during this period is constant. for the next 20 minutes the car moves with a constant speed of 60 mph. At this time the driver of the car applies the brakes and the car declerates to full stop in 1- secounds. the variation of the speed of the car with time is shown in the diagram below. give your answers using 3 significat digits, using correct unit notation and show all of your work.

what is the acceleration during the first 20 secounds?

what is the acceleration during the last 10 secounds?

what is the acceleration during the 20 th 1220 secounds time period?

what is the total distance traveld by the car?

what is the average speed of the car over the whole distance travelled?

Explanation / Answer

since acceleration is constant so let it be a

by newton's law

v=u + at

where a is acceleration, v is final velocity= 26.666 m/s(=60 miles per hour)....u is final velocity =0 and t is the time =20

using this relationship a=v/t= 1.3333 m/s^2

2)similarly using same relationship for last 10 seconds where v=0, u=26.666 (=60 mph),t=10

we get

a=-26.66/10= -2.66m/s^2

3)between t=20 to t=1220 speed is constant....hence acceleration during this period is zero because v=u

4)total distance D travelled by car is area under v-t curve

=area of given trapezium= 0.5(26.66)(1200+1230)

so D=32391.9 m

5)average speed = total distance travelled / total time taken =32391.9/1230= 26.33 m/s m

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