Am I setting up this kinematic equation properly?

In summary, the car goes a quarter mile in 12.1 seconds, and if you hit the brakes, it would take about 1507.8 meters to stop.
  • #1
TaylorHoward21
10
1
A race car starts from rest and goes a quarter mile (1/4 mi) in 12.1 seconds. Assume the acceleration of the car is constant. (a.) What is the acceleration of the car? (b.) What is the final speed of the car? (c.) If you hit the brakes, how far would it take to stop the car if the breaks cause an acceleration of -2.00 m/s2?

I believe I have parts a and b right, however I am not confident with my process of solving part C.


1/4 mi ≈ 402.3 m

a.) acceleration = 5.5m/s^2
b.) Final speed (at 12.1 seconds) = 66.5 m/s

c.) For part C, I reset the known values to:
a = -2.00m/s2
V0 = 66.5m/s
V = 0 (because solving for car stop which is zero velocity)
X0 = 402.3m

The equation I used: v2 - V20 = 2a(x - x0)
The answer I got was 1507.8m = x
For a car going 66.5m/s that is slowing down at -2.00 m/s2 does this answer seem practical? I am struggling to check my work here.

Many thanks.
 
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  • #2
TaylorHoward21 said:
X0 = 402.3m
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
 
  • #3
haruspex said:
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
So if I change X0 to be = 0 the new answer I get is 1105.5m. This still seems rather far for a car to stop from 66.5m/s to 0m/s. Is -2.00m/s^2 simply just a very slow breaking acceleration?
 
  • #4
TaylorHoward21 said:
Is -2.00m/s^2 simply just a very slow breaking acceleration?
It is rather modest. Safe but serious braking is about 0.5g on a dry road. Emergency braking could go as high as 0.7-1g, depending on the vehicle.

And it is "braking", using "brakes", not "breaking".
 
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1. What is a kinematic equation?

A kinematic equation is a mathematical formula that describes the motion of an object, taking into account its position, velocity, and acceleration.

2. How many kinematic equations are there?

There are five kinematic equations, also known as the "Big Five" equations. They are commonly used to solve problems involving motion and are derived from the three basic kinematic equations: v = u + at, s = ut + 1/2at^2, and v^2 = u^2 + 2as.

3. What are the variables in a kinematic equation?

The variables in a kinematic equation are: s (position), u (initial velocity), v (final velocity), a (acceleration), and t (time). These variables may also be represented by different symbols, such as x, y, and z for position, and i, j, and k for velocity and acceleration in vector notation.

4. How do I know if I am setting up the kinematic equation properly?

To set up a kinematic equation properly, you should first identify the known and unknown variables in the problem. Then, choose the appropriate equation that only includes those variables. Make sure to pay attention to the direction of motion and use proper units. Finally, plug in the values and solve for the unknown variable.

5. Can I use kinematic equations for any type of motion?

Kinematic equations are applicable for any type of motion, whether it is linear, circular, or projectile motion, as long as the acceleration is constant. For non-constant acceleration, more advanced equations, such as the calculus-based equations of motion, may be used.

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