Acceleration of a Truck Problem

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The truck accelerates from rest at 2.00 m/s² until it reaches 20.0 m/s, taking 10 seconds. It then travels at this constant speed for 20 seconds before decelerating to a stop over 5 seconds. The total time the truck is in motion is 10 seconds of acceleration, 20 seconds of constant speed, and 5 seconds of deceleration, totaling 35 seconds. The average velocity can be calculated by dividing the total distance traveled by the total time. The calculations need to account for all phases of motion to determine the correct average velocity.
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Homework Statement


A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck travels for 20.s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
a) How long is the truck in motion?
b) What is the average velocity of the truck for the motion described?

Homework Equations


Constant acceleration equations
average speed formula

The Attempt at a Solution


vfinal=20
vinitial=0
t=?
a=2.00

vfinal=vinitial + at
20=2t
t=10s

vaverage=\frac{xfinal=xinitial}{tfinal-tinitial}
=\frac{100}{10}
=10m/s

Now this is incorrect but can someone tell me the correct way of doing this problem?
 
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Paymemoney said:

Homework Statement


A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck travels for 20.s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
a) How long is the truck in motion?
b) What is the average velocity of the truck for the motion described?

Homework Equations


Constant acceleration equations
average speed formula

The Attempt at a Solution


vfinal=20
vinitial=0
t=?
a=2.00

vfinal=vinitial + at
20=2t
t=10s

vaverage=\frac{xfinal=xinitial}{tfinal-tinitial}
=\frac{100}{10}
=10m/s

Now this is incorrect but can someone tell me the correct way of doing this problem?

Are you saying that the truck was in motion for only 10 sec.? How can that be, since the problem states that the truck accelerates for a while, coasts along at a constant speed for 20 seconds, and then brakes to a stop, taking another 5 sec.
 
can someone tell me if the information given in the following statement has any impact on the calculations and if so how would i use it.
Then the truck travels for 20.s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
 
Paymemoney said:
a) How long is the truck in motion?
b) What is the average velocity of the truck for the motion described?
The total time the truck is in motion is the time it is accelerating plus the time it is moving at a constant speed plus the time it is decelerating to a stop.

The average velocity = (total distance)/(total time)
 
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