How Do You Calculate Stopping Time and Deceleration for a Car?

[davidavado]

Homework Statement

A car is traveling along a highway with a speed of 25m/s when the driver sees an obstruction 180m directly ahead. It takes the driver 0.80s to react and begin breaking.

A) How long will it take the car to stop once brakes are applies, provided the car stops just before the obstruction?

B) What is the value of acceleration of the car just before hitting the obstruction? Assume acceleration is uniform.

Homework Equations

d= v (t)

a = v₂-v₁ / t₂ - t₁

d = v₁(t) + 1/2 a (t²)

The Attempt at a Solution

I don't understand how or know how to use two equations with missing variables to find a variable.

 

[Seannation]

Well the car is going at 25m/s whilst the driver is taking time to react, so you can work out the distance the car goes before the brakes are applied and take this distance away from 180m. This new number is the distance you need to work out the braking time for.

Now you know the distance the car takes to brake, and the change in velocity (25m/s to 0m/s). Do you know a way to work out the acceleration required to stop the car?

[tex]v^2 = u^2 + 2as[/tex]

Where v = final velocity, u = initial velocity, a = acceleration and s = distance. Rearrange this equation to get acceleration.

Once you have the acceleration (it should be a negative number, since velocity is decreasing with time), you can use your acceleration equation to work out the time taken. Remember to add on the 0.8s taken for the driver to react.

 

[kerimek]

I would suggest breaking down the problem into smaller parts and using the given information to solve for the missing variables.

A) To find the time it takes for the car to stop, we can use the equation d = v1(t) + 1/2 a (t2). We know the initial velocity (v1) is 25m/s and the final velocity (v2) is 0m/s since the car stops. We also know the distance (d) is 180m. We can rearrange the equation to solve for time (t): t = (2d/v1)^1/2. Plugging in the values, we get t = (2*180m/25m/s)^1/2 = 6.74s. Therefore, it will take the car 6.74 seconds to stop once the brakes are applied.

B) To find the acceleration of the car, we can use the equation a = v2-v1 / t2-t1. We already know the final velocity (v2) is 0m/s and the initial velocity (v1) is 25m/s. We also just calculated the time (t) to be 6.74s. We can rearrange the equation to solve for acceleration (a): a = (v2-v1) / (t2-t1). Plugging in the values, we get a = (0m/s - 25m/s) / (6.74s - 0.80s) = -3.31m/s^2. The negative sign indicates that the car is decelerating. Therefore, the value of acceleration just before hitting the obstruction is -3.31m/s^2.