Two moving cars (Kinematics problem)

[Adjoint]

Homework Statement

Two cars, A and B are traveling in same direction with velocities v_A and v_B. When car A is distance d behind car B, the breaks on A are applied, causing a deceleration a. Show that to prevent a collision between A and B, it is necessary that, v_A - v_B < $\sqrt{2ad}$

Homework Equations

1D kinematics

The Attempt at a Solution

Suppose at the moment car A starts acceleration, it is at the origin, then car B is at x = d.
After time t, car A has position v_At - $\frac{1}{2}$at²
Then car B has the position v_Bt + d

Now, to avoid collision, v_Bt + d > v_At - $\frac{1}{2}$at²
→ v_A - v_B < $\frac{1}{2}$at + $\frac{d}{t}$

Now I have to remove t from this equation.
Any help please?

Thanks in advance.

 

[adjacent]

Two cars, A and B are traveling in same direction with velocities v_A and v_B. When car A is distance d behind car B, the breaks on A are applied, causing a deceleration a. Show that to prevent a collision between A and B, it is necessary that, v_A - v_B < $\sqrt{2ad}$

If the car A is behind the car B, then how can it cause a collision when the brakes of A are applied?

 

[Adjoint]

If the car A is behind the car B, then how can it cause a collision when the brakes of A are applied?

Well, I think there can be a collision if car A was moving with a much higher velocity than car B. So even after being behind and applying break, it might hit car B.

 

[jbriggs444]

Try to solve for the time at which the collision will occur.

 

[Adjoint]

v_Bt + d > v_At - $\frac{1}{2}$at²
→ v_A - v_B < $\frac{1}{2}$at + $\frac{d}{t}$

I can solve the time of collision by taking this inequality as an equality. Is that what you mean?

 

[dauto]

I can solve the time of collision by taking this inequality as an equality. Is that what you mean?

Yes, solve for the time of collision and figure out what condition must be satisfied in order for a solution to exist. If a solution does not exist then the collision doesn't happen.

 

[adjacent]

Well, I think there can be a collision if car A was moving with a much higher velocity than car B. So even after being behind and applying break, it might hit car B.

Oh, that. :shy:
Yeah, it is possible. :shy:

 

[Adjoint]

Yes, solve for the time of collision and figure out what condition must be satisfied in order for a solution to exist. If a solution does not exist than the collision doesn't happen.

Thanks!

We have, v_Bt + d = v_At - $\frac{1}{2}$at²
→ $\frac{1}{2}$at² + (v_B - v_A)t + d = 0
Solving for t,
t = $\frac{-(v_B - v_A) \pm \sqrt{(v_B - v_A)^2 - 4.\frac{1}{2}ad}}{2.\frac{1}{2}at^2}$
If we want a solution not to exist then (v_B - v_A)² - 2ad < 0
→ (v_B - v_A)² < 2ad

Now we know if x ≤ y then √x ≤ √y if both x, y ≥ 0
As there is a chance of two cars to collide we must assume v_A > v_B

So to take square root on both side we rearrange the equation as
(v_A - v_B)² < 2ad
→ v_A - v_B < √(2ad)

 

[Adjoint]

Oh, that. :shy:
Yeah, it is possible. :shy:

At first, I also thought how can there be a collision?
But later I also had this Oh! I see moment. :tongue2:

 

[jbriggs444]

Good job with the solution. Note that there is another approach that is arguably easier.

Nothing in the situation depends on the speed of the cars relative to the road. All that matters is their velocity, distance and acceleration relative to each other.

So make your job easy -- assume that the front car is stationary. Now the equation is simple.

 

[Adjoint]

I hadn't thought this way but now I see that can be another approach. Thank you.