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30. A truck on a straight road starts from rest and accelerates at 2.9 m/s? until it reaches a speed of 32 m/s. Then the truck travels for 33 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 20.0 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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Answer #1


In this question we are going to use the following relations to solve all the parts where s is the distance traveled, v is the final velocity, u is the initial velocity and a is the acceleration We will first break the motion into three phases, first phase when the truck is moving at a constant acceleration, then second phase is the one where truck moves at constant velocity and the third is the one where truck applies its brakes First Phase for the first phase of the motion we have u 0, v 32, a 2.9 using equation (2) we will get 32 0+ 2.9t solving for t we will get, t 11.0345s In this phase the total distance traveled by the truck will be, using equation (1) s0 x 11.03452.9 x 11.03452 176.552nm Second Phase after this in the second phase of the motion we have, u 32, a 0, t 33

the distance traveled in this phase is sut -32 x 33 1056m Third Phase now in the third phase of the motion, u 32, t 20, v 0 using equation (2) we will get 0- 32+a x 20 solving for a we will get, a 1.6 from this a we can find the distance traveled in this phase as x 1.6x 202 320m - total time taken by the truck for the entire journey will be 11.0345 33 + 20- 64.0345 To find the average velocity we will need to find the total distance covered by the truck. Which is given by 176.552 +1056 320-1552.55m (10) the average velocity will be given by Total displacement 1552.55 Time Taken 64.0345 24.24m/s

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