It takes 4.58 seconds for the car to stop and the initial speed of the car is 77.14 ft/s
Explanation:
Given:
Acceleration, a = -17 ft/s²
Distance, s = 175 ft
Final velocity, v = 0
Time, t = ?
Initial speed, u = ?
We know:
v² - u² = 2as
(0)² - (u)² = 2 X - 17 X 175
u² = 5950
u = 77.14 ft/s
To calculate time:
v = u + at
0 = 77.14 - 17 X t
17t = 77.14
t = 4.58 sec
Therefore, It takes 4.58 seconds for the car to stop and the initial speed of the car is 77.14 ft/s
After the driver of a car slams on the brakes, which causes the car to stop after traveling 175 ft with a slow down rate of 17 ft/s², we have:
a) The car stops after 4.54 s.
b) The car's initial speed is 77.14 ft/s.
We can calculate the time at which the car stops with the following kinematic equation:
[tex] v_{f} = v_{i} + at [/tex] (1)
Where:
t: is the time =?
[tex] v_{i}[/tex]: is the initial velocity
[tex] v_{f}[/tex]: is the final velocity = 0 (the car stops)
a: is the acceleration = -17 ft/s² (the minus sign is because the car is slowing down its speed)
First, we need to find the initial velocity.
[tex]v_{i}^{2} = v_{f}^{2} - 2ad[/tex]
Where:
d: is the distance traveled = 175 ft
Then, the initial velocity is:
[tex]v_{i} = \sqrt{v_{f}^{2} - 2ad} = \sqrt{-2(-17 ft/s^{2})175 ft} = 77.14 ft/s[/tex]
Hence, the initial velocity is 77.14 ft/s.
Now, the time at which the car stops is (eq 1):
[tex] t = \frac{v_{f} - v_{i}}{a} = \frac{0 - 77.14 ft/s}{-17 ft/s^{2}} = 4.54 s [/tex]
Therefore, the car stops after 4.54 s.
Find more about acceleration here:
I hope it helps you!
