A speeder passes a police car while traveling 130km/h. the police car imediently starts to accelerate at 10m/s^2 until the police car catches up to the speeder. how long does it take the police car to catch up?
First, convert the speed to m/s
Vcar = 130 km/h = 36.11 m/s
Solve for t:
36.11*t = (a/2) t^2 = 5 t^2
where a is the acceleration rate of the poice car.
Ignore the t=0 solution. Take the other one.
To find the time it takes for the police car to catch up to the speeder, we can use the equation of motion for constant acceleration. The equation is:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this case, the speeder's velocity is 130 km/h, which needs to be converted to m/s.
To convert km/h to m/s, we divide by 3.6 (since 1 km/h = 1000 m/3600 s = 1/3.6 m/s). So, the speeder's velocity is:
u = 130 km/h ÷ 3.6 = 36.11 m/s (rounded to two decimal places)
The police car starts from rest, so the initial velocity (u) is 0 m/s.
The acceleration (a) of the police car is given as 10 m/s².
Using the equation of motion:
v = u + at
We can rearrange the equation to solve for time (t):
t = (v - u) / a
Plugging in the known values:
t = (36.11 m/s - 0 m/s) / 10 m/s²
Simplifying:
t = 36.11 s / 10
t = 3.61 seconds (rounded to two decimal places)
Therefore, it would take the police car approximately 3.61 seconds to catch up to the speeder.