A truck on a straight road starts from rest, accelerating at 2 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s. (i) Plot the position-time, velocity-time and acceleration-time graphs for this motion.
Answers
Assumption: to find the average speed we can break all path up into three phases:
accelerating → coasting → decelerating.
First phase (accelerating)
Rate equation for the acceleration phase:
V1 = a1t1
t1 = V1/ a1 = 20 m s /2.0 m s^2
= 10 s
It means thet accelerating will stop after 10 seconds because then the truck will reach velocity 20 m/s.
Equation of motion for the acceleration phase:
S1 = a1t1 2 /2
= 2.0 ∙ (10 s)^2 /2
= 100 m
Second phase (coasting):
Truck travels path at 20 m/s for 20 s:
S2 = V2t2 = 20 ∙ 20s = 400 m
Third phase (decelerating):
Rate equation for the acceleration phase:
0 = V2 − a3t3
a3 = V2 /t3
= 20 /5
= 4
It means thet accelerating will stop after 10 seconds because then the truck will reach velocity 20 m/s.
Equation of motion for the deceleration phase:
S3 = a3t3 /2
=
4 ∙ (5 s) /2
= 50 m
Average velocity = all distance /all time
= S1 + S2 + S3 /t1 + t2 + t3
100m + 400m + 50m /10s + 20s + 5s =
= 15.7
Time of the travel is = 10s + 20s + 5s = 35s.
average velocity during the motion is 15.7 ,truck was in motion during 35s.