# Kinematics - Find position

1. Oct 25, 2016

I've made a drawing in order to visualize the problem better:

1. The problem statement, all variables and given/known data

A car can increase its speed only at 3.00 m/s2, move at constant velocity, or decrease its speed at 7.00 m/s2. Starting at rest, the driver wishes to drive to a road sign located at x=100 m. He increases his speed, then travels at constant velocity and then decreasing his speed.
The driver begins accelerating at t=0 and switches to constant velocity at t=4.50 s. At what position must the driver switch from constant velocity to decreasing speed if he wants to stop the car at the road sign?

2. Relevant equations
These are the kinematic equations I used to solve the problem:
x1=x0+v0(t1-t0) + 1/2a0(t1-t0)2

v1=v0+a0(t1-t0)

v32=v22 + 2a2(x3-x2)

3. The attempt at a solution
x1=x0+v0(t1-t0) + 1/2a0(t1-t0)2
x1=1/2a0t12 =
x1=1/2(3.0m/s2)(4.50s)2 = 30.4m

v1=v0+a0(t1-t0)
v1=(3.0m/s2)(4.50s) = 13.5m/s

Since after v1 the velocity is constant, v2 must be constant and equal to 13.5m/s

v32=v22 + 2a2(x3-x2)
-v22=2a2x3-2a2x2
x2=-[-(v22-2a2x3) / 2a2]
x2 = -[(-182.25+1400)/-14] = 86.98m

Does this result look right?

2. Oct 25, 2016

### PeroK

Can you think of a way to check? Perhaps you may see an easier way to do it.

3. Oct 25, 2016

I only see that I can skip the first step in finding x1, but I can't think of a way to check or find x2 in a different way. Any hints?

4. Oct 25, 2016

### PeroK

Work backwards. Think about stopping distance for the car.

5. Oct 25, 2016

Do you mean to check if x2 is equal to 86.98m or to find x2 in another way? The only way to check if x2 is equal to 86.98m is to use the same formula I used to find this value which is:
v32=v22+ 2a2(x3-x2)
Since we don't have any values for t3 or t2

Also, in order to find x2 working backwards, I think I still need to find v2 first. I'm a little confused. Perhaps I'm missing some relationship in between time and constant velocity.

6. Oct 25, 2016

### PeroK

Suppose the car is moving along at $13.5 ms^{-1}$. It has a fixed deceleration of $7ms^{-2}$. From that you can calculate the stopping distance of the car. Once you know the stopping distance, you can subtract that from $100m$.

7. Oct 25, 2016

The only way I can think of based on the formulas I learned would be subtracting x2-x1. So it would be 86.98m-30.4m=50.6m
However, I don't understand this approach because based on my sketch I uploaded on the first post, I set x2 as the position in which the driver must switch from constant velocity to decreasing speed if he wants to stop the car at the road sign.

8. Oct 25, 2016

### PeroK

A car is moving along at $13.5 ms^{-1}$. It has a fixed deceleration of $7ms^{-2}$.

a) How long does it take to stop (time)?
b) How far does it travel before it stops?

Can you work that out?

9. Oct 25, 2016

Yes.
a)
v1=v0+a0(t1-t0)
t1=1.93 s

b)
v12=v02+2a0(x1-x0)
-182.25=-14x1
x1=13.02 m

10. Oct 26, 2016

### PeroK

So, you need to start decelerating $13.02m$ before the point you want to stop. And what is $100m - 13.02m$?

11. Oct 26, 2016

So 100m - 13.02m = 86.98 m would be the position in which the driver must switch from constant velocity to decelerating in order to stop the car at 100m correct?

12. Oct 26, 2016

### PeroK

Yes, exactly. And now you have a much simpler solution to the original problem.

13. Oct 26, 2016