# Acceleration and velocity question-easy-need help

## Homework Statement

A car is waiting at a stoplight. Just as the light turns green and the car starts to accelerate, a truck passes at a constant velocity in the next lane. The truck passes at a velocity 25.0 m/s while the car accelerates 5.00 m/s^2 both in the same direction,
Calculate the distance travelled by the car before it overtakes the truck?
Car's speed as it overtakes the truck?

## Homework Equations

I am a little confused on this one.

## The Attempt at a Solution

I think you have to use the formula vf^2=Vi^2 +2ad but am not sure. Do you combine 2 formulas?

any help would be appreciated

## Homework Statement

A car is waiting at a stoplight. Just as the light turns green and the car starts to accelerate, a truck passes at a constant velocity in the next lane. The truck passes at a velocity 25.0 m/s while the car accelerates 5.00 m/s^2 both in the same direction,
Calculate the distance travelled by the car before it overtakes the truck?
Set the position of the truck and car equal to one another and solve for time. Then use the time to solve for the position again
Car's speed as it overtakes the truck?
From the first section you now have all you need to solve for Vx. The magnitude of Vx is Speed.

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1. Homework Statement

Set the position of the truck and car equal to one another and solve for time. Then use the time to solve for the position again
Car's speed as it overtakes the truck?
From the first section you now have all you need to solve for Vx. The magnitude of Vx is Speed.
.............................

When I solve for t it become zero, therefore the distance becomes zero which is impossible. any other suggestions

When I solve for t it become zero, therefore the distance becomes zero which is impossible. any other suggestions

You should probably show how you solved for time, because you clearly made an error at some point.

Recheck your equation. This is how it goes:

X = Xi + Vi(t) + (.5)Ax(t^2)
Xc = 0 + 0(t) + (.5)(5.00)(t^2) = 2.5t^2
XT = 0 + 25.0(t) + (.5)(0)(t^2) = 25.0t

SO....2.5t^2 = 25.0t.......THEN.....t=10.

You should probably show how you solved for time, because you clearly made an error at some point.

car 1-

vi=0 m/s
a=5.00 m/s^2

Truck

v=25.0 m/s

assuming the distances are the same at the beginning
distance would equal zero as they are stopped

using the formula d=Vi(t)+1/2at^2

0=0t+1/2(5.00)t^2
0=1/2(5.00)t^2
0=5.00t^2
0=t^2
t=0

plug that back into the formula

D=Vf(t)-1/2at^2
D=25(t)-1/2(5.00)t^2
D=0

this is my work for this and it is not getting me the right answer which in the back of the book says 250m for A and 50 m/s for B

car 1-

vi=0 m/s
a=5.00 m/s^2

Truck

v=25.0 m/s

assuming the distances are the same at the beginning
distance would equal zero as they are stopped

They are definitely not stopped considering the truck has a constant velocity.

The reason we set the distance equal to one another (NOT TO 0, because we dont know the distance) is that we are finding WHEN the car meets up with the trucks position. By solving for t, as I did above, you do that.

this is my work for this and it is not getting me the right answer which in the back of the book says 250m for A and 50 m/s for B

So if you look at my equation above and then plug in t=10 to either X function.....you get exactly 250m.

Also, the reason you were getting t=0 is that you were solving for when the cars were at distance 0...which is t = 0.

Also, the reason you were getting t=0 is that you were solving for when the cars were at distance 0...which is t = 0.

I am still a bit confused with the x's u are using in your formula. We were given the formula

d=vi(t) + 1/2at^2

how did u get all of those x's into it and 0's for the x's

thanks for all the help!

I am still a bit confused with the x's u are using in your formula. We were given the formula

d=vi(t) + 1/2at^2

how did u get all of those x's into it and 0's for the x's

thanks for all the help!

So the whole equation is:

d(the position of x from Xi aka the distance) = Xi(Initial position of x which is 0 and why they probably left it out) + Vi(t)(which is initial velocity, 0 for the car cause it is at rest in the beginning, and 25 is given for the truck) + 1/2Axt^2(I put the x because it just means Acceleration in the x direction, but you probably haven't went into acceleration in the y direction yet so that is why it is tripping you up. Just think that Ax is equal to what you have for a, Also, A is 0 for the truck because the velocity is constant, and 5.0 is given for the car.)

That should cover everything, hope that helps.

So the whole equation is:

d(the position of x from Xi aka the distance) = Xi(Initial position of x which is 0 and why they probably left it out) + Vi(t)(which is initial velocity, 0 for the car cause it is at rest in the beginning, and 25 is given for the truck) + 1/2Axt^2(I put the x because it just means Acceleration in the x direction, but you probably haven't went into acceleration in the y direction yet so that is why it is tripping you up. Just think that Ax is equal to what you have for a, Also, A is 0 for the truck because the velocity is constant, and 5.0 is given for the car.)

That should cover everything, hope that helps.