A driver in a car traveling at a speed of
76 km/h sees a cat 109 m away on the road.
How long will it take for the car to acceler-
ate uniformly to a stop in exactly 106 m?
Answer in units of s
the average velocity during deceleration is __ (76 km/h + 0) / 2
the time is the distance divided by the average velocity
you will have to convert km/h to m/s to find the answer in s
17M/S2
To find the time it takes for the car to accelerate uniformly to a stop, we need to use the equations of motion. The two equations we'll be using are the equation of motion with uniform acceleration:
Vf^2 = Vi^2 + 2ad
and the equation:
Vf = Vi + at
where:
Vf is the final velocity (which is 0, since the car comes to a stop)
Vi is the initial velocity (given as 76 km/h)
a is the acceleration (unknown)
d is the distance (given as 106 m)
t is the time (unknown)
Let's solve these equations step by step to find the answer:
First, we need to convert the initial velocity from km/h to m/s since the distance is given in meters. We divide 76 km/h by 3.6 (since 1 km/h = 1/3.6 m/s) to get:
Vi = 76 km/h ÷ 3.6 = 21.11 m/s (rounded to two decimal places)
Now, using the equation Vf = Vi + at and since the final velocity is 0, we can rearrange the equation to solve for time (t):
0 = 21.11 m/s + at
Next, we use the equation Vf^2 = Vi^2 + 2ad. We know Vf = 0, Vi = 21.11 m/s, and d = 106 m. Solving for acceleration (a):
0 = 21.11^2 + 2a(106)
0 = 445.0321 + 212a
-445.0321 = 212a
a = -445.0321 / 212
a = -2.1002 m/s^2 (approximated to four decimal places)
Substituting the value of acceleration (a) into the equation 0 = 21.11 + at:
0 = 21.11 + (-2.1002)t
-21.11 = -2.1002t
t = -21.11 / -2.1002
t = 10.05 seconds (rounded to two decimal places)
Therefore, it will take approximately 10.05 seconds for the car to accelerate uniformly to a stop in exactly 106 meters.