how long will it take for a car to accelerate to come to a stop if it is traveling at 20 m/s and the car has a maximum deceleration of 6 m/s squared?
So far I think im supposed to do
6-20/26
Think of the units:
(20 m/s) / (6 m/s^2) = 3.33 s
Think of it like this. Every second the speed decreases by 6 m/s. So, it takes 3.33 seconds to lose all 20 m/s.
Or, how about this:
a = ∆v/∆t
so, ∆t = ∆v/a = (0-20)/-6 = = 3.33
To calculate the time it will take for the car to come to a stop, you need to use one of the kinematic equations of motion. In this case, you can use the equation:
v = u + at
Where:
v = final velocity (0 m/s in this case because the car comes to a stop)
u = initial velocity (20 m/s)
a = acceleration (deceleration in this case, which is -6 m/s²)
t = time
Rearranging the equation to solve for time (t), you get:
t = (v - u) / a
Plugging in the values from your problem:
t = (0 - 20) / -6
Simplifying the equation:
t = -20 / -6
t = 3.33 seconds (rounded to two decimal places)
Therefore, it will take approximately 3.33 seconds for the car to come to a stop.