A "rocket car" is launched along a long straight track at = 0 . It moves with constant acceleration = 2.5 . At = 3.0 , a second car is launched with constant acceleration = 8.7. At what time does the second car catch up with the first one? How far have the cars traveled when the second passes the first?
A car starts from rest and travels for 4.2 s with a uniform acceleration of +1.9 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.6 m/s2. The breaks are applied for 1.60 s. How fast is the car going at the end of the braking period?
A sailboat starts from rest and accelerates at a rate of 0.19 m/s2 over a distance of 384 m.Find the magnitude of the boat's final velocity.
To find the time at which the second car catches up with the first car, we need to equate their positions. Let's denote the position of the first car as x1 and the position of the second car as x2.
The position of an object can be determined using the following equation of motion:
x = x0 + v0t + (1/2)at^2
where x is the position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.
For the first car, we have:
x1 = x1_0 + v1_0t + (1/2)a1t^2
Since the initial velocity (v1_0) is given as 0 and the acceleration (a1) is given as 2.5, the equation becomes:
x1 = x1_0 + (1/2)(2.5)t^2
Similarly, for the second car, we have:
x2 = x2_0 + (1/2)(8.7)t^2
Since both cars start from the same initial position (x1_0 = x2_0 = 0), we can equate the positions of the two cars:
x1 = x2
x1 = x2_0 + (1/2)(8.7)t^2 [Substituting x2 into x1]
x2_0 + (1/2)(8.7)t^2 = (1/2)(2.5)t^2
Simplifying the equation:
4.35t^2 = 8.7t^2
4.35t^2 - 8.7t^2 = 0
-4.35t^2 = 0
Since both t^2 cannot be negative, the only solution is t = 0.
Therefore, the second car catches up with the first car at t = 0, which means they start at the same time.
To find the distance traveled by both cars when the second car passes the first, we can substitute the value of t = 3.0 into either of the equations for x1 or x2:
For x1:
x1 = (1/2)(2.5)(3.0)^2
= 11.25
So, the first car has traveled a distance of 11.25 units when the second car passes it.
For x2:
x2 = (1/2)(8.7)(3.0)^2
= 39.15
So, the second car has traveled a distance of 39.15 units when it passes the first car.
Therefore, the second car catches up with the first car at t = 0, and the first car has traveled 11.25 units while the second car has traveled 39.15 units when this happens.