Respuesta :
Answer:
option (C)
Explanation:
Speed of car = 30 m/s
Let the time taken by the police car to catch the speeding car is t
The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t
Distance traveled by the police car in time t
[tex]s=ut + 0.5 at^{2}[/tex] .... (1)
Distance traveled by the speeding car in t + 1 second
s = 30 (t + 1) = 300
t + 1 = 10
t = 9 s
Put the value of t in equation (1), we get
300 = 0 + 0.5 x a x 9 x 9
a = 7.41 m/s^2
C. 7.41 meters per square second.
In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:
Car
[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)
Police car
[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)
Where:
- [tex]x_{o}[/tex] - Initial position, in meters.
- [tex]x_{C}[/tex] - Final position of the car, in meters.
- [tex]x_{P}[/tex] - Final position of the car, in meters.
- [tex]t[/tex] - Time, in seconds.
- [tex]t'[/tex] - Delay time, in seconds.
- [tex]a_{P}[/tex] - Acceleration of the police car, in meters per square seconds.
- [tex]v_{C}[/tex] - Speed of the car, in meters per second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:
[tex]300 = 30\cdot t[/tex] (1)
[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)
By (1):
[tex]t = 10[/tex]
Then we find that acceleration of the police car must be:
[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]
[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]
Therefore, the correct choice is C.
We kindly invite to check this question on kinematics: https://brainly.com/question/24544574
