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.
Find the speed of car 1 just before it collides with car 2.
Note: The correct answer does not depend on the variable: t
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 Vo. Car 1 begins to move at t=0, speeding up with a constant acceleration Ax. Car 2 continues to move with a constant velocity.
Find the speed of car 1 just before it collides with car 2.
D = the distance between them at t = 0.
V2 = the constant velocity of car 2.
A = the constant acceleration of car 2.
x = the time of travel of car 2 to impact with catr 1.
t = the time of travel to impact.
1--t = x/V2 or x = tV2
2--D - x = At^2/2
3--D - x + x = D = At^2/2 + tV2
4--At^2/2 + tV2 - D = 0
5--At^2 + 2tV2 - 2D = 0
6--Using the quadratic equation,
---t = [sqrt(V2^2 + 2DA)- V2]/a
7--From Vf = Vo + At, the final velocity of car 1 at impact is
---Vf1 = sqrt(V2^2 + 2DA) - V2
Example:
D = 1000 miles
V2 = 100 miles/hr
A = 10 miles/hr^2
After a few guesses, x = 732 miles from which t = 7.32 hr, from which (D - x) = 268 miles which should equal AT^2/2 = 10(7.32)^2/2 = 268 miles and Vf = At = 73.2 mph.
Using Vf1 = sqrt(V2^2) + 2DA0) - V2,
Vf1 = sqrt((100^2) 2(1000)10 - 100 = 73.2mph.
Therefore, Vf1 = sqrt(V2^2 + 2DA) - V2
To find the speed of Car 1 just before it collides with Car 2, we need to determine the time it takes for them to collide. Since the correct answer does not depend on the variable 't', we can solve this problem using kinematic equations.
Let's denote the final speed of Car 1 just before the collision as v_1. The speed of Car 2 remains constant at v_0.
Step 1: Determine the time it takes for Car 1 to collide with Car 2.
Using the equation of motion s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken:
For Car 1:
s_1 = 0.5at^2, since the initial velocity is 0.
For Car 2:
s_2 = v_0 * t, since Car 2 is moving at a constant velocity.
At the time of collision, the distances traveled by the two cars should be equal:
0.5at^2 = v_0 * t
Simplifying the equation:
0.5at^2 - v_0 * t = 0
Step 2: Solve the quadratic equation.
Rearranging the equation:
0.5at^2 - v_0 * t = 0
=> t * (0.5at - v_0) = 0
Since the time taken cannot be zero, we have:
0.5at - v_0 = 0
=> 0.5at = v_0
=> t = (2 * v_0) / a
Step 3: Find the speed of Car 1 just before the collision.
To find the speed, we can use the equation of motion v = u + at, where u is the initial velocity, a is the acceleration, and t is the time taken.
For Car 1:
v_1 = 0 + a * t
= a * (2 * v_0) / a
= 2 * v_0
Therefore, the speed of Car 1 just before colliding with Car 2 is 2 * v_0.
To find the speed of car 1 just before it collides with car 2, we need to determine the time it takes for the two cars to collide. Once we have the time, we can calculate the speed of car 1 at that time.
First, let's find the time it takes for the two cars to collide. We can use the fact that the relative displacement between the two cars is decreasing at a constant rate. Taking the positive direction as the direction towards car 2, the relative displacement is given by:
x_rel = D - v_0*t
where x_rel represents the relative displacement, D is the initial distance between the cars, v_0 is the initial velocity of car 2, and t is the time.
Now, let's consider the motion of car 1. It starts from rest and accelerates with a constant acceleration a_x. We can use the equation of motion for car 1 to relate time and distance:
x_1 = (1/2)*a_x*t^2
where x_1 represents the displacement of car 1.
At the moment of collision, the displacement of car 1 must equal the relative displacement, so we can set x_1 = x_rel:
(1/2)*a_x*t^2 = D - v_0*t
Rearranging the equation, we get a quadratic equation:
(1/2)*a_x*t^2 + v_0*t - D = 0
Now, we can solve this equation for t using the quadratic formula:
t = (-v_0 ± sqrt(v_0^2 + 2*a_x*D)) / a_x
Since we're only interested in the positive solution, we take the plus sign in the formula.
Now that we have the time t, we can calculate the speed of car 1 just before collision. The final speed of car 1 is given by the equation of motion:
v_1 = 0 + a_x*t
Substituting the value of t, we find:
v_1 = a_x * [(-v_0 + sqrt(v_0^2 + 2*a_x*D)) / a_x]
The acceleration a_x cancels, resulting in:
v_1 = -v_0 + sqrt(v_0^2 + 2*a_x*D)
Therefore, the speed of car 1 just before colliding with car 2 is given by the above expression.