A driver in a car traveling at a speed of
66 km/h sees a cat 118 m away on the road.
How long will it take for the car to accelerate uniformly to a stop in exactly 114 m?
Answer in units of s
66 km/h=66000/3600=18.33 m/s
a=v²/2s=18.33²/2•114=1.47 m/s²
v=at
t=v/a=18.33/1,47=12.43 s
To find the time it takes for the car to accelerate uniformly to a stop, we can use the equations of motion.
First, we need to determine the initial velocity of the car. The driver sees the cat 118 m away, which means the car needs to travel a total distance of 118 m + 114 m = 232 m before coming to a stop.
The initial velocity, u, is given as 66 km/h. To convert it to m/s, we need to divide by 3.6 since there are 3.6 seconds in one hour.
So, u = 66 km/h ÷ 3.6 = 18.33 m/s.
Now, we need to find the acceleration, a. We know that the car is accelerating uniformly to a stop, so its final velocity, v, is 0 m/s. We also know the initial velocity, u, and the displacement, s, which is 232 m.
We can use the equation: v^2 = u^2 + 2as, where v = 0 m/s.
0 = (18.33 m/s)^2 + 2a(232 m)
Rearranging the equation, we get:
0 - (18.33 m/s)^2 = 2a(232 m)
-335.4 m^2/s^2 = 464a
Now, we can solve for the acceleration, a:
a = -335.4 m^2/s^2 ÷ 464 m
a ≈ -0.723 m/s^2
Finally, we can use the equation of motion v = u + at to find the time, t, it takes for the car to come to a stop.
0 = 18.33 m/s + (-0.723 m/s^2) * t
Rearranging the equation, we get:
-18.33 m/s = -0.723 m/s^2 * t
t = -18.33 m/s ÷ -0.723 m/s^2
t ≈ 25.37 s
Therefore, it will take approximately 25.37 seconds for the car to accelerate uniformly to a stop in exactly 114 m.