A "rocket car" is launched along a long straight track at t=0s. It moves with constant acceleration a1=3.0m/s^2. At t=2.5s, a second car is launched with constant acceleration a2=9.0m/s^2.
1. At what time does the second car catch up with the first one?
2. How far have the cars traveled when the second passes the first?
1/2 * 3t^2 = 1/2 * 9 (t-2.5)^2
t = 5.9
plug in that value for t into either expression.
1. To find the time when the second car catches up with the first car, we can calculate the time it takes for the second car to cover the same distance as the first car. Since both cars move with constant acceleration, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity
t = time taken
a = acceleration
Let's start by finding the distance traveled by the first car at time t = 2.5s. Using the equation for the first car:
s1 = (1/2)a1t1^2
s1 = (1/2)(3.0 m/s^2)(2.5s)^2
s1 = (1/2)(3.0 m/s^2)(6.25 s^2)
s1 = (0.5 m/s^2)(6.25 m^2/s^2)
s1 = 3.125 m
Now, let's find the time it takes for the second car to cover the same distance. Using the equation for the second car:
s2 = (1/2)a2t2^2
s2 = (1/2)(9.0 m/s^2)(t2)^2
Since the second car starts at time t = 2.5s, we can subtract that time from the total time taken:
s2 = (1/2)(9.0 m/s^2)(t2 - 2.5s)^2
To find the time when the second car catches up with the first car, we set s2 equal to s1:
(1/2)(9.0 m/s^2)(t2 - 2.5s)^2 = 3.125 m
Simplifying:
(9.0 m/s^2)(t2 - 2.5s)^2 = 6.25 m
Now, we can solve for t2 - 2.5s:
(t2 - 2.5s)^2 = 6.25 m / (9.0 m/s^2)
(t2 - 2.5s)^2 = 0.694 s^2
Taking the square root of both sides:
t2 - 2.5s = ±√(0.694) s
t2 - 2.5s = ±0.833 s
Now we have two possible values for t2:
Case 1: t2 - 2.5s = 0.833 s
t2 = 0.833 s + 2.5s
t2 = 3.333 s
Case 2: t2 - 2.5s = -0.833 s (discard this case since time cannot be negative)
Therefore, the second car catches up with the first car at t = 3.333 seconds.
2. To find how far the cars have traveled when the second car passes the first, we can find the distance traveled by the second car at t = 3.333s using the equation of motion:
s2 = (1/2)a2t2^2
s2 = (1/2)(9.0 m/s^2)(3.333 s)^2
s2 = (1/2)(9.0 m/s^2)(11.1105 s^2)
s2 = (0.5 m/s^2)(11.1105 m^2/s^2)
s2 = 5.55525 m
Therefore, the cars have traveled a distance of 5.55525 meters when the second car passes the first.
To find the time at which the second car catches up with the first car and the distance traveled by both cars at that time, we can use the equations of motion. Let's solve each question separately:
1. To find the time at which the second car catches up with the first car, we need to determine when they are at the same position. We can use the equation of motion to find the position of each car at any given time:
For the first car (car 1):
s1 = u1*t + 0.5*a1*t^2
For the second car (car 2):
s2 = u2*t + 0.5*a2*t^2
Where s1 and s2 are the distances traveled by car 1 and car 2, u1 and u2 are the initial velocities of car 1 and car 2, a1 and a2 are the accelerations of car 1 and car 2, and t is the time.
Since car 1 starts earlier than car 2, its initial velocity is 0. Therefore, u1 = 0. Car 2 starts at t = 2.5s, so its initial velocity is:
u2 = a2*(2.5s)
To find the time at which the second car catches up with the first car, we need to solve the equation s1 = s2:
0.5*a1*t^2 = u2*t + 0.5*a2*t^2
Simplifying the equation:
0.5*a1*t^2 - 0.5*a2*t^2 - u2*t = 0
This is a quadratic equation in t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/(2a),
where a = 0.5*(a1 - a2), b = -u2, and c = 0.
Plug in the known values and solve for t. You will get two values: one positive and one negative. Since time cannot be negative, the positive value will give us the time at which the second car catches up with the first car.
2. To find the distance traveled by both cars when the second car passes the first, substitute the value of t obtained from the previous step into the equation of motion for either car to calculate the distance traveled by both cars.
s1 = u1*t + 0.5*a1*t^2
or
s2 = u2*t + 0.5*a2*t^2
You can use either equation to find the distances traveled by both cars at the time t.
Plug in the known values (including the value of t obtained in the previous step) and solve for the distances traveled by both cars.