A car with good tires on a dry road can decelerate at about 5.00m/s^2 when braking. If the car is traveling at 34.0m/s, how long does it take the car to stop under these conditions?
Not quite sure if this is the correct procedure, but what I would have done is use the formula:
a=[delta]v/t since we are given a and t.
so 5=34/t and I got t=6.8s.
Where did you find this question? Just out of curiosity.
Vf=Vi+at
solve for time t.
**Havanna**The question was from a test that I had taken and I had gotten it wrong--I was just curious on which equation to use. And it turns out to be as simple as I thought it might be. Thank you both for your help!
To find out how long it takes for the car to stop, we need to use the equation of motion. The equation we can use is:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (which is 0 since the car comes to a stop)
vi = initial velocity (34.0 m/s)
a = acceleration (deceleration in this case, which is -5.00 m/s^2)
d = displacement (unknown)
Let's rearrange the equation to solve for d:
d = (vf^2 - vi^2) / (2a)
Substituting the known values into the equation:
d = (0 - (34.0 m/s)^2) / (2 * -5.00 m/s^2)
Simplifying the equation:
d = (-1156.0 m^2/s^2) / -10.0 m/s^2
d = 115.6 m
Now that we know the displacement, we can use it to find the time taken using the formula:
t = (vf - vi) / a
Substituting the known values into the equation:
t = (0 - 34.0 m/s) / -5.00 m/s^2
Simplifying the equation:
t = (-34.0 m/s) / -5.00 m/s^2
t = 6.80 s
Therefore, it takes approximately 6.80 seconds for the car to stop under the given conditions.