A boy on a motor scooter is traveling at a velocity of 14.1 m/s when he applies his
brakes in order to stop. How much time will it take him to stop if he slows down
at a rate of 3.2 m/s2?
Well, well, well, looks like this boy is trying to put on a grand show of braking! Bravo! Now, let's get into the calculations before you put on your own show of impatience.
We can use the equation of motion to find the time it takes for the boy to stop. The equation goes like this:
v = u + at
where:
v = final velocity (which is 0 m/s because he wants to stop, duh!)
u = initial velocity (14.1 m/s)
a = acceleration (in this case, it's the deceleration because he wants to stop, even more duh!)
t = time (the thing we want to find)
Plug in the given values, and we get:
0 = 14.1 + (-3.2)t
Now, let's solve this equation for t. Drum roll, please!
-14.1 = -3.2t
Divide both sides by -3.2, and voilà, we have our answer!
t = 4.40625 seconds
So, it will take approximately 4.41 seconds for our braking boy to come to a halt. Happy stopping, my friend!
To find the time it will take for the boy to stop, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s, since he is stopping)
u = initial velocity (14.1 m/s)
a = acceleration (slowing down at a rate of 3.2 m/s^2)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 14.1) / (-3.2)
Simplifying the equation:
t ≈ -14.1 / -3.2
t ≈ 4.41 seconds
Therefore, it will take the boy approximately 4.41 seconds to stop.
To find the time it takes for the boy to stop, we can use the equation of motion:
v = u + at
where:
v = final velocity (which is 0 m/s since he wants to stop)
u = initial velocity of the scooter (14.1 m/s)
a = acceleration (negative because it's slowing down, -3.2 m/s^2 in this case)
t = time
Rearranging the equation to solve for time:
t = (v - u) / a
Substituting the given values:
t = (0 - 14.1) / -3.2
t = 14.1 / 3.2
t ≈ 4.41 seconds
Therefore, it will take him approximately 4.41 seconds to stop.
0 = 14.1 - 3.2 t
as Janet told you below.