# Determine the car's displacement

1. Oct 7, 2007

### etSahcs12

Hi, I am new to the fourm. If it is not any touble, please help me with a kinematic problem.

1. The problem statement, all variables and given/known data

A car travels east along a straight road at a constant velocity of 18 m/s. After 5.0s, it accelerates uniformly for 4.0s, reaching a velocity of 24 m/s. For the next 6.0s, the car proceeds with uniform motion. Determine the car's displacement for the 15.0s trip.

2. Relevant equations

Δd = v1 x Δt + 1/2 x acceleration x Δt(squared)
Δd = v2 x Δt - 1/2 x acceleration x Δt(squared)
Δd = 1/2 (v1 + v2) Δt
2aΔd = v2^2 - v1^2
a = v2 - v2 / Δt

3. The attempt at a solution

This is the only question I have not experienced before. Please guide me through this question.

Last edited: Oct 7, 2007
2. Oct 7, 2007

### Kurdt

Staff Emeritus
You will have to split the problem into 3 parts. There are two where the car is travelling with a constant velocity and one where it is accelerating. The equation you have stated is almost correct. For constant velocity you will need:

$$d = vt$$

For constant acceleration you will need one of the constant acceleration kinematic equations.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

3. Oct 7, 2007

### etSahcs12

so do i do:

18 m/s x 5.0s then 24 x 4.0s then add them up?

What do i do with the 6.0s and the uniform motion? When adding do i find the displacement?

4. Oct 7, 2007

### Kurdt

Staff Emeritus
No. The two time intervals when the car is travelling with constant velocity are the first 5 seconds and the last 6 seconds after the acceleration. To work out the distance travelled while accelerating you will need one of the kinematic equations in the link I gave you above. To save you the time here it is:

$$d = ut +\frac{1}{2} at^2$$

5. Oct 7, 2007

### etSahcs12

How am i able to find the acceleration and v1 by using all of these variables (specifically time and velocity)? Would i use $$d = vt$$ at all?

Last edited: Oct 7, 2007
6. Oct 7, 2007

### HallsofIvy

Staff Emeritus
Which question are you talking about? You have posted on question and were told to break it into 3 parts.
Okay, how far will it go at 18 m/s for 5 s?

Hint: at uniform acceleration, the "average speed" is just the average of the lowest and highest speeds: (18+24)/2 m/s. How far will the car go in 4.0s at that average speed?

At 24 m/s how far will the car go in 6.0 s? NOW add them all up.

7. Oct 7, 2007

### etSahcs12

Ohh i see. If i were to add them, I would get the displacement correct? or should i use $$d = ut +\frac{1}{2} at^2$$ after adding.

8. Oct 7, 2007

### Kurdt

Staff Emeritus
You can ignore that one, I made a mistake. You will need to use the average velocity as HallsofIvy suggested in d = vt.

9. Oct 7, 2007

### etSahcs12

so, bastically i do $$d = vt$$ with the time and velocities, add them up and thats my displacement?

Last edited: Oct 7, 2007
10. Oct 7, 2007

### fanj06

A rifle is aimed horizontally at a target located 44.0 m away. The bullet hits the target 3.0 cm below the aim point. What was the bullet's time of flight?

11. Oct 7, 2007

### Kurdt

Staff Emeritus
Yes but remember the middle time stint will require you work out the average speed.

12. Oct 7, 2007

### etSahcs12

I see, so after adding 24+18 and dividing by 2, multply the time and that will be my distance ? Once done, do the same for the other 2 (but without finding the average acceleration) and add, thus giving my displacement?

Last edited: Oct 7, 2007
13. Oct 7, 2007

### Kurdt

Staff Emeritus
Yes.