A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 29.0 m/s. Then the truck travels for 25.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?

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fichoh fichoh · Answer 1 · 2021-01-25T05:39:37+01:00

Answer:

44.5s ; 22.64 m/s

Explanation:

The motion of the truck could be separated into 3 different phases :

First :

Time of motion :

Initial Velocity, u = 0 ; final velocity, v = 29 m/s

Acceleration, a = 2 m/s²

Recall: acceleration = change in velocity / time

Time = change in velocity / acceleration

Time = (29 - 0) / 2 = 14.5 second

Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m

Second :

Time = 25 seconds at constant speed

29 m/s for 25 seconds

v*t = 29 * 25 = 725 m

Third:

5 seconds before coming to rest

((29 + 0) /2) * 5

14.5 * 5 = 72.5 m

A.)

Length of journey = (14.5+ 25 + 5) = 44.5 seconds

B.)

Average velocity = total distance / total time taken

Average velocity = (210.25 + 725 + 72.5) / 44.5

= 1007.75 / 44.5

= 22.646067

= 22.64 m/s