Calculate Acceleration of Car: 1600m in 0.125s

[ProBasket]

a car starts from rest and acclerates down a straight track of length L= 1600m with a constant accleration. If the time it takes for the car to travel the final d= 100 m of the track (from 1500m to 1600m) is T=0.125s, then the acceleration of the car is... the answer is 80.7 m/s^2

this isn't hw, it's actually practice for the final i have tommorow. I haven't done this question in awhile, so I am not sure how to do it. but here's what i think i should do... I am having trouble finding which equation to use. well i have the initial x and final x, and t. and i need to find acceleration. so i think i need to use this formula:

$$X = x_0 + v_0*t + 1/2at^2$$

100 = 1500 + 1/2at^2

x final should be 100 right? because the final 100 it needs to travel
and solve for a? doesn't really make sense. can someone offer some help on which equation i should use?

 

[OlderDan]

You know you are starting from rest, so you can take

$$x = x_0 + v_0*t + \frac{1}{2}at^2 = 0 + 0 + \frac{1}{2}at^2$$

You know two distances for x (1500, 1600), and you know the time difference between them. If t is the first time, then t + .125s is the second time. Put your times into the equation with the known distances and you will have two equations for two unknowns that you can solve for a and t.

 

[thepatientmental]

To solve for acceleration, you can use the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity (which is 0 in this case since the car starts from rest), and t is the time taken. In this scenario, vf is the velocity at the end of the 100m track, which can be calculated by dividing the distance (100m) by the time (0.125s). So vf = 100m/0.125s = 800m/s. Plugging in these values into the formula, we get a = (800m/s - 0m/s)/0.125s = 6400m/s^2. However, this is the average acceleration over the entire 1600m track. To find the acceleration for just the final 100m, we can use the formula a = 2(xf - xi)/t^2, where xf is the final position (1600m), xi is the initial position (1500m), and t is the time taken (0.125s). Plugging in these values, we get a = 2(1600m - 1500m)/(0.125s)^2 = 80.7m/s^2. Therefore, the acceleration of the car for just the final 100m of the track is 80.7m/s^2.