# Homework Help: Drag Racing Acceleration Question

1. Feb 26, 2016

### Grace204

1. The problem statement, all variables and given/known data
Drag racing usually involves two cars racing each other over a set distance. Although distances range from 200 m to 1 km, the 400.0 m drag race is the most popular. This race tests a vehicle's acceleration and top speed.
Data collected on a race by a dragster-
Distance (m) | 20.0 | 400.0 |
Time (s) | 0.945| 8.96 |

The acceleration of the dragster, from the 20.0 m mark to the 400.0 m mark, is?
2. Relevant equations
v= d/t
a= (Vf - Vi)/ t
3. The attempt at a solution
Vi= d/t
= 20.0m/0.945s
= 21.16m/s
Vf= d/t
= 400.0m/8.96s
= 44.64 m/s
a = (Vf- Vi)/t
= (44.64 - 21.16)/ (8.96 - 0.945)
= 2.93 m/s2

I know that this isn't right because the answer is supposed to be 1.27 m/s2. But I can't figure out how to solve it to get this answer. If anyone could tell me if I'm using the wrong formula or doing something wrong that would be great.

2. Feb 26, 2016

### cnh1995

This means there is no acceleration from 0m to 20m mark..Is that given in the problem?
This is a sudden change in speed( in zero time) and no acceleration from 20 to 400m mark..
I believe the car has a constant acceleration a1 from 0 to 20m and a2 from 20 to 400 m.

3. Feb 26, 2016

### Grace204

I don't really understand what you mean by that; could you explain further?

4. Feb 26, 2016

### rcgldr

Distance (m) | 20.0 | 400.0 |
Time (s) | 0.945| 8.96

So a distance of 380 (480-20) meters is run in 8.015 (8.960 - 0.945) seconds, for an average speed of 47.41 m/s. The issue is the acceleration and speed at 20 meters is unknown. For the given answer of 1.27 m / s^2, the increase in speed over 8.015 seconds is only 10.18 m/s, meaning the speed at 20 meters would have to be 37.23 m/s, which seems unlikely.

5. Feb 26, 2016

### cnh1995

From 0 to 20m, you calculated speed as 21.16m/s. This means the car is moving with constant speed from 0 to 20m. So, there is no acceleration. Also, from 20m to 400m, you calculated the speed as 44.64m/s. This means, the car changed its speed from 21.16m/s to 44.64m/s in zero time. This is not possible. Here's what I think of the scenario:
The car starts from rest and has constant acceleration a1 from 0 to 20m span. Then from 20m to 400m span, it has acceleration a2. Is there anything else given in the problem regarding acceleration?

6. Feb 26, 2016

### SammyS

Staff Emeritus
It looks as if what cnh1995 states is what is meant in the problem. Maybe it is and maybe it isn't a very satisfactory approximation to the motion of the drag racing vehicle.

Don't forget, for constant acceleration, $\displaystyle \ v_\text{avg}=\frac{v_f+v_i}{2}\$.

Putting this all together does give a ≈ 1.27 m/s2 .

7. Feb 27, 2016

### Grace204

How does it equal 1.27? I've been trying, but I just can't seem to figure it out.

8. Feb 27, 2016

### SammyS

Staff Emeritus
Let $\ t_1=0.945\text{s}\,,\ x_1=20\text{m, and } v_1 \$be the velocity at $\ t_1\$. Furthermore, let $\ \bar v_1 \$ be the average velocity from time 0 to $\ t_1 \$.

Similarly:
Let $\ t_2=8.96\text{s}\,,\ x_2=400\text{m, and } v_2 \$be the velocity at $\ t_2\$. Furthermore, let $\ \bar v_2 \$ be the average velocity from time $\ t_1 \$ to time $\ t_2 \$.

Find $\ \bar v_1 \$ and $\ \bar v_2 \$. From those, find $\ v_1 \$ and then $\ v_2 \$. Then find average acceleration from time $\ t_1 \$ to time $\ t_2 \$.

*Edit:
There was a typo in the above It's now corrected to read as x2 = 400 m.

Last edited: Feb 27, 2016
9. Feb 27, 2016

### Grace204

Now I understand, except why would x1 be equal to 300 m?

10. Feb 27, 2016

### Grace204

Also, I found that velocity 1 = 21.16 m/s and velocity 2= 33.48 m/s
How do I find v1 and v2 from there? Can you tell me the formula please?

11. Feb 27, 2016

### SammyS

Staff Emeritus
20/0.945 ≈ 21.16 m/s is the average velocity from t = 0 to t1 . Assuming uniform acceleration over that time, what is the instantaneous velocity, v1, at time t1 ?

(See the corrected value for x2 .)