8. A speeding caris 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/S?. O How fast will it be going when it catches up with the speeding car?
To solve this problem, we can use the equation of motion:
Vf = Vi + at
where:
Vf is the final velocity of the police car,
Vi is the initial velocity of the police car (0 m/s as it is stopped),
a is the acceleration of the police car (+7.0 m/s?),
t is the time it takes for the police car to catch up with the speeding car.
Now, let's calculate the time it takes for the police car to catch up with the speeding car. We can use the equation:
Distance = Speed × Time
The distance covered by the speeding car in that time will be the same distance covered by the police car.
Distance = Speed × Time
Distance = 30.0 m/s × t
Distance = (Vi × t) + (1/2 × a × t²) (applying the equation of motion for the police car)
Since the initial velocity of the police car is zero (Vi = 0), the equation becomes:
Distance = 0 × t + (1/2 × 7.0 m/s? × t²)
Distance = 0 + (3.5 m/s? × t²)
Distance = 3.5t²
Now, let's equate the distances covered by both cars:
30.0 m/s × t = 3.5t²
Dividing both sides by t:
30.0 m/s = 3.5t
Solving for t:
t = 30.0 m/s / 3.5
t ≈ 8.57 seconds (rounded to 2 decimal places)
Now that we have the time it takes for the police car to catch up with the speeding car, we can substitute it into the equation of motion to find the final velocity of the police car:
Vf = Vi + at
Vf = 0 m/s + 7.0 m/s? × 8.57 s
Vf ≈ 60.0 m/s
Therefore, the police car will be going at approximately 60.0 m/s 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:
\(v_f = v_i + at\)
where:
\(v_f\) is the final velocity of the police car.
\(v_i\) is the initial velocity of the police car.
\(a\) is the acceleration of the police car.
\(t\) is the time it takes for the police car to catch up.
We know that the initial velocity of the police car (\(v_i\)) is 0 m/s, as it was stopped. The acceleration of the police car (\(a\)) is +7.0 m/s^2.
The speeding car is already traveling at a constant speed of +30.0 m/s, which means its velocity (\(v\)) remains constant.
The time it takes for the police car to catch up with the speeding car is the same for both vehicles. Therefore, we can set the equation for the police car equal to the equation for the speeding car:
\(v_f = v + at\)
Substituting the values we know:
\(v_f = 30.0 \, \text{m/s} + 7.0 \, \text{m/s}^2 \cdot t\)
Since the time it takes for the police car to catch up is the same for both vehicles, we can equate the two equations:
\(30.0 \, \text{m/s} + 7.0 \, \text{m/s} \cdot t = v + 7.0 \, \text{m/s} \cdot t\)
Simplifying the equation, we find:
\(30.0 \, \text{m/s} = v\)
Therefore, the police car will be going at a speed of 30.0 m/s when it catches up with the speeding car.