A speeding motorists passes a stopped police car. At the moment he passes, the police car begins accelerating at a constant rate of 4.4 m/s^2. The motorist unaware that he is being chased, continues at a constant speed until the police car catches him 12s later. How fast was the motorist going?
d1 = d2.
V1*t1 = 0.5a*t^2.
V1*12 = 0.5*4.4*12^2.
Solve for V1(m/s).
Well, you see, the motorist was probably going at quite the speed if he managed to catch the attention of the police car! But of course, the police car had to speed up in order to catch up to him. And after some calculating (or perhaps some clown car logic), it seems the motorist was going approximately 52.8 m/s. That's definitely fast enough to make the police car's siren go "wee-oo-wee-oo"! Just remember, speed limits are no laughing matter, my friend. Stay safe on the roads!
To solve this problem, we can use the equations of motion. Let's call the initial velocity of the motorist "u", the acceleration of the police car "a", the time taken "t", and the final velocity "v".
Since the motorist continues at a constant speed, his final velocity (v) is equal to his initial velocity (u). And since the police car starts from rest, its initial velocity is 0.
Using the equation of motion: v = u + at
Substituting the known values, we have:
u + 4.4t = u
Simplifying the equation, we have:
4.4t = 0
Dividing both sides of the equation by 4.4, we get:
t = 0
This implies that the motorist was caught instantly by the police car. Therefore, it is likely that there is an error in the given information or the problem is incomplete.
To solve this problem, we can use kinematic equations to find the speed of the motorist.
Let's consider the motion of the police car first. We know that it begins accelerating at a constant rate of 4.4 m/s^2 after the motorist passes. Since the police car catches up with the motorist in 12 seconds, we can apply the equation of motion:
d = vi*t + (1/2) * a * t^2
Where:
d is the distance traveled
vi is the initial velocity
a is the acceleration
t is the time
In this case, the distance traveled by the police car is equal to the distance traveled by the motorist.
Now let's denote the speed of the motorist as v_m, and the initial speed of the police car as v_p.
Initially, the motorist is ahead of the police car, so the initial distance between them is the sum of their initial speeds multiplied by the time taken by the police car to catch up. Therefore:
d = (v_m - v_p) * t
Since we know the police car's acceleration is 4.4 m/s^2, we can substitute the value of d from the previous equation into the equation of motion:
(v_m - v_p) * t = (1/2) * 4.4 * t^2
Now, rearranging the equation:
v_m - v_p = 2.2 * t
Since the acceleration of the motorist is zero (constant speed), the motorist's final speed is equal to the initial speed, v_m = v_f.
Therefore:
v_f - v_p = 2.2 * t
Substituting the value of t as 12 seconds:
v_f - v_p = 2.2 * 12
v_f - v_p = 26.4
Finally, since the motorist's final speed is equal to the initial speed, we have:
v_p = v_f - 26.4
The initial velocity of the police car, v_p, is unknown. However, we are interested in the speed of the motorist, so we don't need to solve for v_p.
Hence, the speed of the motorist at the time of passing the police car is 26.4 m/s.