8. A speeding car is traveling at a constant speed of +30.0 m/s when it
passes a stopped police car. The police car accelerates at +7.0 m/s2.0
How fast will it be going when it catches up with the speeding car?
To determine how fast the police car will be going when it catches up with the speeding car, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity of the police car
u = initial velocity of the police car
a = acceleration of the police car
s = displacement
In this scenario, the initial velocity of the police car (u) is 0 m/s (since it is stopped), the acceleration (a) is +7.0 m/s^2, and the displacement (s) will be the same for both vehicles since they will meet at some point.
To find the displacement, we can use the equation:
s = ut + (1/2)at^2
Where:
t = time taken for the police car to catch up with the speeding car
Since the speeding car is traveling at a constant speed, the displacement s will be equal to the product of the constant speed and the time taken (t) for the police car to catch up.
Therefore, s = 30.0 m/s * t
Now we can substitute this value of s into the first equation to find the final velocity (v) of the police car:
v^2 = 0^2 + 2 * 7.0 m/s^2 * (30.0 m/s * t)
Simplifying,
v^2 = 2 * 7.0 m/s^2 * (30.0 m/s * t)
v^2 = 420 m/s * t
Taking the square root of both sides,
v = √(420 m/s * t)
Since this equation gives the final velocity (v) of the police car in terms of time (t), we need to know the time it takes for the police car to catch up with the speeding car in order to find the final velocity.
Unfortunately, this information is not provided, so we cannot calculate the exact final velocity of the police car.
To find out how fast the police car will be going when it catches up with the speeding car, we can use the equation of motion:
v² = u² + 2as
where:
v = final velocity (unknown),
u = initial velocity of the police car (0 m/s),
a = acceleration of the police car (+7.0 m/s²),
s = distance traveled by the police car.
Since the speeding car is traveling at a constant speed, the distance traveled by the police car will be the same as the distance traveled by the speeding car.
Let's assume the distance traveled by both cars is "d."
For the speeding car:
u = +30.0 m/s
a = 0 m/s² (since it is traveling at a constant speed)
s = d
Substituting the values into the equation, we get:
v² = (+30.0 m/s)² + 2(0 m/s²)(d)
v² = 900 m²/s²
For the police car:
u = 0 m/s
a = +7.0 m/s²
s = d
Substituting the values into the equation, we get:
v² = (0 m/s)² + 2(+7.0 m/s²)(d)
v² = 14d m²/s²
Since the distance traveled by both cars is the same ("d"), we can equate the two equations:
900 m²/s² = 14d m²/s²
Solving for "d," we get:
d = 900 m²/s² / 14 m²/s²
d ≈ 64.29 m
Now we can substitute the value of "d" back into either equation to find the final velocity of the police car.
Using the equation for the police car:
v² = 14d m²/s²
v² = 14(64.29 m)
v² ≈ 900 m²/s²
Taking the square root of both sides to solve for "v," we get:
v ≈ √900 m/s
v ≈ 30.0 m/s
Therefore, the police car will be traveling at approximately 30.0 m/s when it catches up with the speeding car.