Acceleration and velocity question-easy-

In summary, the conversation involves solving for the distance traveled by a car accelerating at 5.00 m/s^2 before overtaking a truck moving at 25.0 m/s in the same direction. The solution involves setting the position of the car and truck equal to each other and solving for time using the formula d=Vi(t)+1/2at^2, then using that time to solve for the position and speed of the car. The distance traveled by the car was found to be 250 m and the speed when overtaking the truck was 50 m/s.
  • #1
vicsic
13
0

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 traveled 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
 
Physics news on Phys.org
  • #2
vicsic said:

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 traveled 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.

.......
 
  • #3
USN2ENG said:
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
 
  • #4
vicsic said:
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.
 
  • #5
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.
 
  • #6
Jokerhelper said:
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

any help please.
 
  • #7
vicsic said:
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 don't 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

any help please.

So if you look at my equation above and then plug in t=10 to either X function...you get exactly 250m.
 
  • #8
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.
 
  • #9
USN2ENG said:
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!
 
  • #10
vicsic said:
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.
 
  • #11
USN2ENG said:
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.

thanks i got the answer!

:)
 

1. What is the difference between acceleration and velocity?

Acceleration is the rate of change of velocity over time, while velocity is the rate of change of position over time. In simpler terms, acceleration is how quickly an object's speed is changing, while velocity is how fast it is moving in a specific direction.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between positive and negative acceleration?

Positive acceleration, also known as acceleration in the positive direction, is when an object's velocity increases over time. Negative acceleration, or acceleration in the negative direction, is when an object's velocity decreases over time.

4. How does acceleration affect an object's motion?

Acceleration can change the speed, direction, or both of an object's motion. When an object experiences acceleration, it will either speed up, slow down, or change direction depending on the direction of the acceleration.

5. What is the difference between average acceleration and instantaneous acceleration?

Average acceleration is the overall change in velocity divided by the total time, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration gives a general idea of an object's motion, while instantaneous acceleration provides a more detailed understanding of how the object is moving at a specific point in time.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
625
  • Introductory Physics Homework Help
Replies
2
Views
848
  • Introductory Physics Homework Help
Replies
5
Views
644
  • Introductory Physics Homework Help
Replies
5
Views
974
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
697
Back
Top