A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, how long does it take to catch the speeder?
they travel the same distance, in the same time
change 100km/hr to 27.8m/s
ds=27.8*t
dp=1/2 3.6*t^2
set them equal
27.8 t= 1.8 t^2 solve for t.
why are you multiplying by 1/2
with constant acceleration a:
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
here Vi = Xi = 0
so
x = (1/2)at^2
how do you find the distance it has to travel before it catches speeder?
Well, it sounds like the police car has quite the need for speed! Let's see if it can catch that speedy speeder.
First, let's convert the speed of the speeder into meters per second. 100 km/h is about 27.78 m/s. Now, we can start the race!
To catch up to the speeder, the police car needs to accelerate from 0 m/s to 27.78 m/s. We know that the police car can accelerate at a rate of 3.6 m/s^2.
Using a little physics magic called the equation of motion, v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time), we can rearrange the equation to solve for time.
Plugging in all the numbers, we have:
27.78 = 0 + (3.6)t
Simplifying the equation, we get:
27.78 = 3.6t
Dividing both sides of the equation by 3.6, we find:
t = 27.78 / 3.6
Calculating that out, we find that it takes approximately 7.72 seconds for the police car to catch the speeder.
So, the police car will need a little over 7.7 seconds to say hello to Mr. Speeder and put an end to their "speedy" endeavors.
To answer this question, we need to calculate the time it takes for the police car to accelerate to the speed of the speeder.
First, let's convert the speed of the speeder to meters per second (m/s). The speed of the speeder is 100 km/h, which is equivalent to 100,000 meters per hour. To convert this to meters per second, we divide by 3,600 (since there are 3,600 seconds in an hour):
Speed of the speeder = 100,000 m / (3,600 s) = 27.78 m/s
Now we can calculate the time it takes for the police car to accelerate to this speed.
The acceleration of the police car is given as 3.6 m/s^2. We can use the following kinematic equation to find the time it takes:
v = u + at
Where:
v = final velocity (speed of the speeder)
u = initial velocity (speed of the police car)
a = acceleration
t = time
In this case, the final velocity (v) is 27.78 m/s and the initial velocity (u) is 0 m/s (since the police car is initially stopped). Plugging in these values into the equation, we get:
27.78 m/s = 0 m/s + (3.6 m/s^2)t
Simplifying the equation:
27.78 m/s = 3.6 m/s^2 * t
Now, solve for t by dividing both sides of the equation by 3.6 m/s^2:
t = 27.78 m/s / 3.6 m/s^2
t ≈ 7.7167 seconds
Therefore, it will take approximately 7.7167 seconds for the police car to catch the speeder.