if a car accelerates from rest at a constant 5.5 m/s^2, how long will it take for the car to reach a velocity of 28 m/s
vo= 0
a=5.5m/s^2
v= 28 m/s
t=?
v=vo+at
at=v-vo
t=(v-vo)/a
t=28/5.5
t=5.09 seconds
Vi= 0
Vf= 28 m/s
a=5.5 m/s²
t=?
Vf=Vi+at
t=Vf+Vi/a
(Since Vi is 0 it can be crossed out)
=28 m/s/5.5 m/s²
=5.09 (m/s/m/s²)
=5.09 m/s
IJOLE, 10 YEARS AGO
Oh, we've got a speedy car here! Let's calculate how long it'll take for it to reach that velocity of 28 m/s.
Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can plug in the values:
28 m/s = 0 m/s + (5.5 m/s^2) * t
Now, let's solve for t:
28 = 5.5t
Dividing both sides by 5.5:
t ≈ 5.09 seconds
So, it'll take approximately 5.09 seconds for the car to reach a velocity of 28 m/s. Just enough time for a quick snack break!
To find the time it takes for the car to reach a velocity of 28 m/s, we can use the basic kinematic equation:
v = u + at
where:
v = final velocity (28 m/s)
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration (5.5 m/s^2)
t = time
Rearranging the equation to solve for time (t):
t = (v - u) / a
Now, substituting the given values:
t = (28 m/s - 0 m/s) / 5.5 m/s^2
Simplifying:
t = 28 m/s / 5.5 m/s^2
t ≈ 5.09 seconds
Therefore, it will take approximately 5.09 seconds for the car to reach a velocity of 28 m/s.