A police car is traveling at a velocity of 15.0 m/s due north, when a car zooms by at a constant velocity of 45.0 m/s due north. After a reaction time 0.700 s the policeman begins to pursue the speeder with an acceleration of 3.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?
it's supposed to be time, not distance as answer
To find out how long it takes for the police car to catch up with the speeder, we need to determine the distance the speeder travels during the reaction time and then calculate the time it takes for the police car to cover that distance.
Let's start by finding the distance the speeder travels during the reaction time of 0.700 s. Since the speeder is traveling at a constant velocity of 45.0 m/s for this period, we can use the formula:
Distance = Velocity × Time
During the reaction time, the distance traveled by the speeder is:
Distance = 45.0 m/s × 0.700 s
Now, let's calculate the distance:
Distance = 31.5 m
After the reaction time, the police car starts to pursue the speeder with an acceleration of 3.00 m/s^2. Both vehicles were initially traveling at the same velocity of 15.0 m/s, so the relative velocity between them is:
Relative Velocity = Speeder's Velocity − Police Car's Velocity
Relative Velocity = 45.0 m/s − 15.0 m/s
Relative Velocity = 30.0 m/s
To find the time it takes for the police car to catch up with the speeder, we can use the equation:
Distance = (Relative Velocity × Time) + (0.5 × Acceleration × Time^2)
Since we already know the distance traveled during the reaction time (31.5 m) and the acceleration (3.00 m/s^2), we can rewrite the equation as:
31.5 m = (30.0 m/s × Time) + (0.5 × 3.00 m/s^2 × Time^2)
Simplifying the equation, we get a quadratic equation:
0.5 × 3.00 m/s^2 × Time^2 + 30.0 m/s × Time - 31.5 m = 0
We can solve this quadratic equation using the quadratic formula or by factoring it, but let's use the quadratic formula:
Time = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 0.5 × 3.00 m/s^2, b = 30.0 m/s, and c = -31.5 m. Plugging in these values:
Time = (-(30.0 m/s) ± √((30.0 m/s)^2 - 4 × 0.5 × 3.00 m/s^2 × (-31.5 m))) / (2 × 0.5 × 3.00 m/s^2)
Now we can solve for Time using a calculator or math software. After obtaining the values for Time, we can check which one is the positive solution since the time cannot be negative in this context.
Once we have the final value for Time, we can add the reaction time (0.700 s) to find the total time it takes for the police car to catch up with the speeder.
HOW TO:
formula: ΔX=Vi(t)+ 1/2(a)(t^2)
what we know:
Vi= 45.0 m/s
t= .700 s
a= 3.00 m/s^2
NOW PLUG IN:
ΔX=45(.7)+1/2(3)(.7^2)
ΔX= 32.235 m