1. ## Finding deceleration

I am trying to solve this problem

"The driver of a vehicle, travelling at 50 km/h, applies the brakes 100 m
from a crosswalk. Assuming constant braking deceleration, find the
deceleration required to bring the vehicle to a full stop at the crosswalk"

I do not really understand how to approach this problem because I cannot easily visualize it. In this case I am not being given any function so I do not really know what to do.

From what I know both velocity and acceleration should be 0 at the moment the car stops but I don't know how to use this information.

Any help will be appreciated

2. Originally Posted by samstark
I am trying to solve this problem

"The driver of a vehicle, travelling at 50 km/h, applies the brakes 100 m
from a crosswalk. Assuming constant braking deceleration, find the
deceleration required to bring the vehicle to a full stop at the crosswalk"

I do not really understand how to approach this problem because I cannot easily visualize it. In this case I am not being given any function so I do not really know what to do.

From what I know both velocity and acceleration should be 0 at the moment the car stops but I don't know how to use this information.

Any help will be appreciated
You will need two kinematic equations to solve this. First

$v_f=v_0+at$

and

$x_f=x_0+v_0t+\frac{1}{2}at^2$

Now we know that

$x_0=0 \quad x_f=100 \quad v_0=50 \quad v_f=0$

So we have two equations and two unknowns a and t.

can you finish from here?

Don't forget to make sure that your units are the same.

3. Originally Posted by TheEmptySet
You will need two kinematic equations to solve this. First

$v_f=v_0+at$

and

$x_f=x_0+v_0t+\frac{1}{2}at^2$

Now we know that

$x_0=0 \quad x_f=100 \quad v_0=50 \quad v_f=0$

So we have two equations and two unknowns a and t.

can you finish from here?

Don't forget to make sure that your units are the same.
Just to make sure that I am going on the right track. Are my two equations supposed to be like this?

0 = 50 + at
100 = 0 + 50t + (1/2)at^2

4. No you have mixed the units!