How Do You Calculate Stopping Distance Using a Velocity-Time Graph?

[getfirefox]

Hello everybody, I have a small question I need help with.

Homework Statement

Draw a sketch graph of the following:

"A car which is traveling at 15 ms-1. At t (time) = 0, the driver sees a pedestrian step out into the road. The driver takes 0.7 seconds to apply the brakes, and the brakes produce an acceleration of -6ms-2, bringing the car to a rest."

Use the stetch graph to calculate the stopping distance.

Homework Equations

None.

The Attempt at a Solution

None. Can somebody give me an example of what the graph would look like and how I would calculate the stopping distance. I take it this would be the displacement, or area under the line on the graph.

Thanks everyone.

 

[tiny-tim]

Draw a sketch graph of the following:

"A car which is traveling at 15 ms-1. At t (time) = 0, the driver sees a pedestrian step out into the road. The driver takes 0.7 seconds to apply the brakes, and the brakes produce an acceleration of -6ms-2, bringing the car to a rest."

Use the stetch graph to calculate the stopping distance.
…
Can somebody give me an example of what the graph would look like and how I would calculate the stopping distance. I take it this would be the displacement, or area under the line on the graph.

Hello getfirefox!

You should plot speed against time.

Constant speed is a horizontal line.

Constant deceleration is a straight line sloping down.

And yes, the stopping distance will be the area under the line.

 

[baba89]

Hello,

To help you better understand the concept behind velocity-time graphs, I will first explain the basic principles and then provide an example for your specific problem.

A velocity-time graph shows the relationship between an object's velocity and time. The slope of the line on the graph represents the object's acceleration, with a steeper slope indicating a higher acceleration. The area under the line represents the object's displacement.

Now, for your problem, the initial velocity of the car is 15 ms-1 and it takes 0.7 seconds for the car to come to a rest. This means that the car's final velocity is 0 ms-1. The brakes produce an acceleration of -6ms-2, which means the car is decelerating.

To draw the graph, you can start by plotting the initial velocity of 15 ms-1 at time t=0. Then, you can draw a straight line with a negative slope of -6ms-2, representing the car's deceleration. The line should intersect the x-axis at t=0.7 seconds, indicating that the car has come to a rest at that point.

To calculate the stopping distance, you can use the equation d=vt+1/2at^2, where d is the displacement, v is the initial velocity, a is the acceleration, and t is the time. In this case, since the final velocity is 0, the equation simplifies to d=1/2at^2. Plugging in the values, we get d=1/2(-6)(0.7)^2= 1.47 meters. This is the stopping distance of the car.

I hope this helps you understand velocity-time graphs better. Let me know if you have any further questions. Good luck with your homework!