Acceleration of a Truck Problem

[Paymemoney]

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=[tex]\frac{xfinal=xinitial}{tfinal-tinitial}[/tex]
=[tex]\frac{100}{10}[/tex]
=10m/s

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

 

[Mark44]

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=[tex]\frac{xfinal=xinitial}{tfinal-tinitial}[/tex]
=[tex]\frac{100}{10}[/tex]
=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.

 

[Paymemoney]

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.

 

[Mark44]

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)

 

[DeeNos]

I would approach this problem by first identifying the known and unknown variables and then using the appropriate equations and principles to solve for the unknowns.

Known variables: Initial velocity (vinitial) is 0 m/s, final velocity (vfinal) is 20 m/s, acceleration (a) is 2.00 m/s^2, and time for acceleration (t1) is 10 s. Time for braking (t2) is 5 s.

Unknown variables: Total time in motion (t), and average velocity (vaverage).

a) To find the total time in motion, we can use the equation vfinal = vinitial + at, where t is the time for acceleration (t1). Rearranging the equation, we get t = (vfinal - vinitial)/a = (20 - 0)/2.00 = 10 s. Since the truck travels for an additional 20 s at constant speed, the total time in motion is 10 s + 20 s = 30 s.

b) To find the average velocity, we can use the formula vaverage = (xfinal-xinitial)/(tfinal-tinitial), where xfinal and xinitial are the final and initial positions of the truck respectively. Since the truck starts from rest, xinitial = 0. The final position can be calculated using the equation x = xinitial + vinitialt + 0.5at^2, where x is the distance traveled, vinitial is the initial velocity, t is the time, and a is the acceleration. Substituting the known values, we get xfinal = 0 + 0 + 0.5(2.00)(10)^2 = 100 m. Therefore, vaverage = (100 - 0)/(30 - 0) = 100/30 = 3.33 m/s.

In summary, the truck is in motion for a total of 30 seconds and has an average velocity of 3.33 m/s. It is important to carefully consider the given information and use the appropriate equations and principles to solve for the unknown variables in a scientific manner.