How Do You Calculate Dragster Deceleration Time and Distance?

[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?

Homework Equations

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).

 

[BvU]

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.

 

[Casey Wilson]

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!

 

[BvU]

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)

 

[HalfLostAndSearching]

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.