Kinematics: Solving for Position with Constant Velocity and Acceleration

[Adriano25]

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

1. Homework Statement
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?

Homework 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)

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?

 

[PeroK]

86.98m

Does this result look right?

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

 

[Adriano25]

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

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?

 

[PeroK]

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?

Work backwards. Think about stopping distance for the car.

 

[Adriano25]

Work backwards. Think about stopping distance for the car.

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.

 

[PeroK]

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.

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 $100m$.

 

[Adriano25]

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.

 

[PeroK]

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.

What about this question:

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?

 

[Adriano25]

What about this question:

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?

Yes.
a)
v₁=v₀+a₀(t₁-t₀)
t1=1.93 s

b)
v₁²=v_0²+2a₀(x₁-x₀₎
-182.25=-14x₁
x₁=13.02 m

 

[PeroK]

Yes.
a)
v₁=v₀+a₀(t₁-t₀)
t1=1.93 s

b)
v₁²=v_0²+2a₀(x₁-x₀₎
-182.25=-14x₁
x₁=13.02 m

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

 

[Adriano25]

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

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?

 

[PeroK]

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?

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

 

[Adriano25]

Great. Thanks for your help!