A speeding car is traveling at a constant 30.0 m/s when it passes a stopped police car. The police car accelerates at 7.0 m/s^2. a)how far will the police travel to catch up to the speeder? b) how fast will it be going when it catches up?
To find the distance the police car will travel to catch up to the speeding car, we need to determine the time it takes for the police car to catch up. We can use the formula:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time^2
Since the initial velocity of the police car is 0 m/s (it starts from rest), the formula simplifies to:
Distance = (1/2) × Acceleration × Time^2
Let's solve the equation for time:
Given:
Initial velocity of the speeding car (u) = 30.0 m/s
Acceleration of the police car (a) = 7.0 m/s^2
1. Calculate the time taken by the police car to catch up:
For the police car to catch up to the speeding car, their velocities should be the same.
Using the equation:
Final Velocity of the speeding car (v) = Initial Velocity of the police car (0) + Acceleration × Time
Substituting the values:
30.0 m/s = 0 + 7.0 m/s^2 × Time
Simplifying,
30.0 m/s = 7.0 m/s^2 × Time
Dividing both sides by 7.0 m/s^2, we get:
Time = 30.0 m/s ÷ 7.0 m/s^2
Calculating,
Time = 4.29 seconds (rounded to two decimal places)
Now, we can find the distance the police car will travel by substituting the calculated time into the distance formula:
Distance = (1/2) × Acceleration × Time^2
Distance = (1/2) × 7.0 m/s^2 × (4.29 s)^2
2. Calculate the distance traveled by the police car:
Using the formula,
Distance = (1/2) × 7.0 m/s^2 × (4.29 s)^2
Calculating,
Distance ≈ 63.35 meters (rounded to two decimal places)
So, the police car will travel approximately 63.35 meters to catch up to the speeding car.
To find the speed of the police car when it catches up, we can use the formula:
Final Velocity = Initial Velocity + Acceleration × Time
Substituting the values:
Final Velocity = 0 + 7.0 m/s^2 × 4.29 s
Calculating,
Final Velocity ≈ 30.03 m/s (rounded to two decimal places)
Therefore, the police car will be going approximately 30.03 m/s when it catches up to the speeding car.
To find the distance the police car will travel to catch up to the speeder:
a) We can use the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
First, let's find the time it takes for the police car to catch up to the speeder. The relative velocity between the police car and the speeder will be equal to the difference in their speeds:
relative velocity = speed of the speeder - speed of the police car
relative velocity = 30.0 m/s - 0 m/s = 30.0 m/s
Now, we can find the time it takes to catch up using the formula:
relative velocity = acceleration * time
30.0 m/s = 7.0 m/s^2 * time
time = 30.0 m/s / 7.0 m/s^2 ≈ 4.29 s
Now we can substitute the time value into the distance equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
distance = 0 m/s * 4.29 s + (1/2) * 7.0 m/s^2 * (4.29 s)^2
distance ≈ 61.22 m
Therefore, the police car will travel approximately 61.22 meters to catch up to the speeder.
b) To find the speed of the police car when it catches up to the speeder, we can use the equation:
final velocity = initial velocity + acceleration * time
final velocity = 0 m/s + 7.0 m/s^2 * 4.29 s
final velocity ≈ 30.03 m/s
Therefore, the police car will be going approximately 30.03 m/s when it catches up to the speeder.
vt=at^2/2
t=2v/a=8.6
s=vt=30•8.6=258 m