A car is traveling at v = 16m/s . The driver applies the brakes and the car decelerates at a= -4.0 m/s^2 . What is the stopping distance?
To find the stopping distance, we can use the equation:
stopping distance = (initial velocity)^2 / (2*acceleration)
Given:
Initial velocity (v) = 16 m/s
Acceleration (a) = -4.0 m/s^2
Substituting the values into the equation:
stopping distance = (16 m/s)^2 / (2 * -4.0 m/s^2)
Simplifying:
stopping distance = 256 m^2/s^2 / -8 m/s^2
stopping distance = -32 m^2/s^2
Since distance cannot be negative, it means the car cannot stop in the given scenario. It will keep moving with the deceleration.
To find the stopping distance, we first need to determine the time it takes for the car to come to a stop. We can use the following equation to find the time:
vf = vi + at
Where:
vf is the final velocity (which is 0 m/s since the car comes to a stop)
vi is the initial velocity (given as 16 m/s)
a is the acceleration (given as -4.0 m/s^2)
t is the time we're trying to calculate.
Rearranging the equation, we get:
0 = 16 + (-4.0)t
Simplifying further:
-16 = -4.0t
Dividing both sides by -4.0, we find:
t = 4.0 seconds
Now that we know the time it takes for the car to stop, we can use another equation to determine the stopping distance:
d = vit + (1/2)at^2
Where:
d is the stopping distance (which we're trying to find)
vi is the initial velocity (given as 16 m/s)
t is the time it takes for the car to stop (calculated as 4.0 seconds)
a is the acceleration (given as -4.0 m/s^2)
Plugging in the values, we get:
d = (16)(4.0) + (1/2)(-4.0)(4.0)^2
Simplifying further:
d = 64 - 32
d = 32 meters
Therefore, the stopping distance of the car is 32 meters.
vi = 16
d = vi t - 2t^2
0 = vi - 4t
t = vi/4 = 4
d = 16*4 - 2(4^2)
= 2*16
=32 meters