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.

(1)

(2)

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

or

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:

(2) Using the equation of velocity:

v = v0 + a · t    (v0 = 0)

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.