# Equations for Accelerated Motion 3

## Homework Statement

A drag racer has an acceleration of 3g. What distance is needed for the car to reach a speed of 320 mi./h (143 m/s)?

v^2=vo^2+2aΔx

## The Attempt at a Solution

I decided to use m/s to make it easier.

143 m/s^2=0+2(9.8*3)Δx
20449=58.8Δx
Δx≈347.7 m
Could someone please check this? I did my best in solving it, and I checked it, but I'd like to make sure before I turn it it.

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
How could you check this yourself?

The only sources of mistake here would be ether your arithmetic or the choice of equation.

You should be able to check your own arithmetic by now - so how would you check you got the right equation?

The best way would be to sketch a v-t graph.
In this case it's a triangle - the line goes from 0 to 142m/s in unknown time T - put those points on the graph (just pick some point on the t axis and label it T).

The displacement (delta-x) is the area under the graph and the acceleration is the slope.
You can write equations for both these, keeping T as an unknown.
Now you have two simultaneous equations and two unknowns ... solve for displacement.

Compare.

Dick
Science Advisor
Homework Helper

## Homework Statement

A drag racer has an acceleration of 3g. What distance is needed for the car to reach a speed of 320 mi./h (143 m/s)?

v^2=vo^2+2aΔx

## The Attempt at a Solution

I decided to use m/s to make it easier.

143 m/s^2=0+2(9.8*3)Δx
20449=58.8Δx
Δx≈347.7 m
Could someone please check this? I did my best in solving it, and I checked it, but I'd like to make sure before I turn it it.

Look fine to me. But Simon does have a good point. At some point you have to start figuring out a way to check these sorts of problems on your own.

Thanks to both of you! I did check it by plugging my answer back into my equation, but since I'm still new to the physics world, I like to triple check just to be safe. :-) Simon, thanks for the suggestion to draw a graph, I forgot about that method.