A car traveling at a constant speed of 33.4 m/s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 3.28 m/s2 How long after the trooper starts the chase does he overtake the speeding car
both equations ..
d=vt
d=1/2 a t^2
distances are the same, set the equations equal, solve for t.
@#6687aA
To find out how long after the trooper starts the chase that he overtakes the speeding car, we can use the equations of motion. Let's denote the time it takes for the trooper to catch up to the car as "t".
First, we need to determine the initial velocity of the trooper's car. Since the trooper starts from rest, the initial velocity (u) of the trooper's car is 0 m/s.
Now, we can use the equation of motion to calculate the displacement (s) of the trooper's car during time "t":
s = ut + (1/2)at^2
Here, "a" represents the acceleration of the trooper's car, which is given as 3.28 m/s^2. The displacement (s) of the trooper's car is the same as the distance between the trooper and the car when the chase begins.
Next, we'll calculate the displacement of the speeding car during the same time "t". Since the speeding car is moving at a constant speed of 33.4 m/s, its initial velocity (u) is 33.4 m/s. Therefore, the displacement (S) of the car during time "t" is given by:
S = ut
We know that when the trooper catches up to the car, the displacements of both vehicles will be equal (s = S).
Substituting the equations for s and S, we have:
ut + (1/2)at^2 = ut
Simplifying the equation, we get:
(1/2)at^2 = 0
Since the equation simplifies to 0, we know that the time it takes for the trooper to catch up to the car is t = 0 seconds. This means the trooper catches up to the speeding car immediately after starting the chase.
Therefore, the trooper overtakes the speeding car one second after starting the chase.