Answer:
(1) [tex]t = -\frac{v_0 - \sqrt{v_0^{2} + 2a_xD}}{a_x}[/tex]
(2)[tex] v = -v_0 + \sqrt{v_0^{2} + 2a_xD}[/tex]
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
[tex]t_1 = -\frac{v_0 + \sqrt{v_0^{2} + 2a_xD}}{a_x}[/tex]
or
[tex]t_2 = -\frac{v_0 - \sqrt{v_0^{2} + 2a_xD}}{a_x}[/tex]
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:
[tex]t = -\frac{v_0 - \sqrt{v_0^{2} + 2a_xD}}{a_x}[/tex]
(2) Using the equation of velocity:
v = v0 + a · t (v0 = 0)
[tex]v = a_x*-\frac{v_0 - \sqrt{v_0^{2} + 2a_xD}}{a_x}[/tex]
[tex] v = -(v_0 - \sqrt{v_0^{2} + 2a_xD})[/tex]
[tex] v = -v_0 + \sqrt{v_0^{2} + 2a_xD}[/tex]
