A driver in a car traveling at a speed of 70 km/h sees a cat 129 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 125 m? answer in units of s
change km/h to m/s
Vf^2=Vi^2+2a*125
solve for a.
Then, vf=vi+at solve for t.
To find the time it takes for the car to accelerate uniformly to a stop, we can use the equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity (which is 0 since the car comes to a stop), u is the initial velocity, a is the acceleration, and s is the distance traveled.
In this scenario, the initial velocity is 70 km/h, which needs to be converted to m/s for consistent units. To convert km/h to m/s, we divide by 3.6:
u = 70 km/h = (70/3.6) m/s = 19.44 m/s.
The distance traveled (s) is 125 m, and since the car starts from rest when it first notices the cat, the initial velocity (u) is 0.
Using the equation of motion, we can solve for the acceleration (a):
v^2 = u^2 + 2as,
0 = (19.44)^2 + 2a(125),
0 = 378.06 + 250a.
Simplifying the equation, we get:
250a = -378.06,
a = -378.06/250,
a = -1.51224 m/s^2.
Now, we can use another equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Rearranging the equation to solve for time (t), we get:
t = (v - u)/a.
Since the final velocity (v) is 0 m/s and the initial velocity (u) is 19.44 m/s, we can substitute these values into the equation:
t = (0 - 19.44)/-1.51224,
t = 19.44/1.51224,
t ≈ 12.86 s.
Therefore, it will take approximately 12.86 seconds for the car to accelerate uniformly to a stop in exactly 125 m.