Equations for Accelerated Motion 3

In summary: I will definitely use it in the future!In summary, the drag racer has an acceleration of 3g and to reach a speed of 320 mi./h (143 m/s), it would need to travel a distance of approximately 347.7 m. This can be checked by sketching a v-t graph and using the equations for displacement and acceleration.
  • #1
Medgirl314
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2

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)?

Homework Equations



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.
 
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  • #2
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.
 
  • #3
Medgirl314 said:

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)?

Homework Equations



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.
 
  • #4
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.
 
  • #5


Your solution is correct. The equation you used is the correct one for this problem. However, it would be helpful to show your units in each step of your calculation, as this will make it easier for others to follow your work and check for errors. Additionally, it is important to note that the value of "g" used in this problem is actually 32.2 ft/s^2 or 9.8 m/s^2, not 3g. Overall, your solution is correct and well thought out. Good job!
 

FAQ: Equations for Accelerated Motion 3

1. What is the formula for calculating acceleration?

The formula for calculating acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

2. Can acceleration be negative?

Yes, acceleration can be negative. This means that an object is slowing down or decelerating.

3. How is velocity related to acceleration?

Velocity and acceleration are directly related. An increase in acceleration results in a corresponding increase in velocity, and vice versa.

4. What is the difference between average and instantaneous acceleration?

Average acceleration is calculated over a specific time interval, while instantaneous acceleration is calculated at a specific moment in time.

5. How can I use equations for accelerated motion to solve real-world problems?

Equations for accelerated motion can be used to calculate important values such as velocity, acceleration, and time in real-world situations. This can help in predicting the motion of objects, designing vehicles and structures, and understanding the forces acting on objects.

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