Drag Racing Acceleration Question

[Grace204]

Homework Statement

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?

Homework Equations

v= d/t
a= (Vf - Vi)/ t

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.

 

[cnh1995]

Vi= d/t
= 20.0m/0.945s
= 21.16m/s

This means there is no acceleration from 0m to 20m mark..Is that given in the problem?

Vf= d/t
= 400.0m/8.96s
= 44.64 m/s

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.

 

[Grace204]

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

 

[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.

 

[cnh1995]

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

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?

 

[SammyS]

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?

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 .

 

[Grace204]

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 .

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

 

[SammyS]

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

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.

 

[Grace204]

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

 

[Grace204]

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_1=300\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 \$.

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?

 

[SammyS]

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?

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 .)