# How Do You Calculate Dragster Deceleration Time and Distance?

• Casey Wilson
In summary, the problem involves a dragster accelerating at 8 m/s^2 for 4.6 seconds and then decelerating to a stop in 100m. The formula x = 0 + 1/2at^2 is used to find the distance traveled in the first part, while a second equation relating initial velocity, final velocity, time, and acceleration is needed to solve for the unknown acceleration and time for the deceleration.
Casey Wilson

## Homework Statement

- A Dragster at the starting line accelerates at 8 m/s^2 to the finish line. If it took 4.6 s, how long is the track?
- The Dragster deccelerated to a stop in 100m. How long did it take?

x = 0 + 1/2at^2

## The Attempt at a Solution

The first part of the questions I got x = 84.64m using the above equation.
For the life of me, I cannot figure out how to get anything viable for other formulas, when v or a is not specified for decceleration (read: Part 2).

Hello casey, welcome to PF :)

Same formula, but now the initial speed isn't zero. And x isn't the unknown, because it's a given. The unknown is a. And t of course. So you'll need another equation. Something relating v initial, v final, t and a.

From there (with v final = 0) you'll find t. (two equations with two unknowns)

You'll do fine.

BvU said:
Hello casey, welcome to PF :)

Same formula, but now the initial speed isn't zero. And x isn't the unknown, because it's a given. The unknown is a. From there (with v final = 0) you'll find t. (two equations with two unknowns)

You'll do fine.

I think I see where you are coming from.
Thank you for the quick reply and the welcome.

I missed the second part of your answer. I think I got it figure out. Thank you!

Ah, some PF culture here: you do the work, helpers help. So jot something down and solicit comments/assistance !
You'll need this other equation anyway (it's no big deal, you must have seen it come by already at some point in the lectures/testbook) to determine the speed when the braking starts.

Hint: check out the formulas here (where it says three key variables)

Last edited:

I would like to clarify that the term "deceleration" can be used interchangeably with "negative acceleration." In this case, the dragster is still accelerating, but in the opposite direction (decelerating) towards a stop. The equation for this situation would be v^2 = u^2 + 2as, where v is the final velocity (which is 0 since the dragster comes to a stop), u is the initial velocity (which we don't know), a is the acceleration (which we know is -8 m/s^2), and s is the distance (which is given as 100m). Rearranging the equation, we get u = √(v^2 - 2as). Plugging in the known values, we get u = √(0^2 - 2(-8)(100)) = 40 m/s.

Now, we can use the equation v = u + at to calculate the time it takes for the dragster to decelerate to a stop. Again, v is 0, u is 40 m/s, a is -8 m/s^2, and we are solving for t. Rearranging the equation, we get t = (v-u)/a. Plugging in the known values, we get t = (0-40)/-8 = 5 seconds.

Therefore, it took 5 seconds for the dragster to decelerate to a stop in 100m.

## 1. What is deceleration in 1D movement?

Deceleration in 1D movement refers to the decrease in speed or velocity of an object moving only in one direction.

## 2. How is deceleration calculated?

Deceleration can be calculated by dividing the change in velocity by the time it takes for the change to occur. The unit for deceleration is meters per second squared (m/s^2).

## 3. What causes deceleration in 1D movement?

Deceleration can be caused by various factors such as friction, air resistance, or an opposing force acting on the object in motion.

## 4. How is deceleration different from acceleration?

Deceleration is the opposite of acceleration. While acceleration refers to the increase in speed or velocity, deceleration refers to the decrease in speed or velocity.

## 5. Can an object experience both acceleration and deceleration in 1D movement?

Yes, an object can experience both acceleration and deceleration in 1D movement. For example, a car can accelerate to a certain speed and then decelerate when the brakes are applied.

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