Question

Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2 is moving to the left with speed v_0. Car 1 begins to move at t=0, speeding up with a constant acceleration a_x. Car 2 continues to move with a constant velocity.(1)At what time do the two cars collide?(2)Find the speed of car 1 just before it collides with car 2.

Answer 1

(1)
`t = -\frac{v_0 - \sqrt{v_0^(2) + 2a_xD}}{a_x}`

(2)
`v = -v_0 + \sqrt{v_0^(2) + 2a_xD}`

Step-by-step explanation:

The equations for position and velocity of the accelerated car (car 1) are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time "t".

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time "t".

For the car 2, the position will be:

x = x0 + v · t

(1)When the cars collide their position is the same. Then:

x car 1 = x car 2

Let´s place the origin of the frame of reference at the point where the car 1 is located. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t (x0 of car 1 and v0 = 0)

1/2 · a · t² = x0 + v · t

1/2 · aₓ · t² = D - v₀ · t

0 = -1/2 · aₓ · t² - v₀ · t + D

Let´s solve the quadratic equation using the quadratic formula:

a = -1/2 · aₓ

b = -v₀

c = D

(-b± √(b² - 4 · a · c)) / 2 · a

t = -(v₀±√(v₀² +2 aₓ · D)) / aₓ

`t_1 = -\frac{v_0 + \sqrt{v_0^(2) + 2a_xD}}{a_x}`

or

`t_2 = -\frac{v_0 - \sqrt{v_0^(2) + 2a_xD}}{a_x}`

The positive time will be t₂ because v₀-√(v₀² +2 aₓ · D) can be negative (if √(v₀² +2 aₓ · D) > v₀) that when multiplied by - 1 will give a positive time. In change, v₀+√(v₀² +2 aₓ · D) will always be positive and when multiplied with -1 will be negative.

Then the two cars will collide at:

`t = -\frac{v_0 - \sqrt{v_0^(2) + 2a_xD}}{a_x}`

(2) Using the equation of velocity:

v = v0 + a · t (v0 = 0)

`v = a_x*-\frac{v_0 - \sqrt{v_0^(2) + 2a_xD}}{a_x}`

`v = -(v_0 - \sqrt{v_0^(2) + 2a_xD})`

`v = -v_0 + \sqrt{v_0^(2) + 2a_xD}`