A car starts from rest and travels for 5.6 s
with a uniform deceleration of 1.3 m/s
2
.
What is the final velocity of the car?
Answer in units of m/s
I know the answer for this is -7.28
but I cant figure out how to the second part.
How far does the car travel in this time interval?
Answer in units of m
d = (1/2) a t^2
since they ask only for distance, a magnitude, and not final position, a vector, you do not have to give a negative sign for the distance.
d = (1/2) (1.3)(5.6)^2
= 20.4 meters
To find the final velocity of the car, we can use the equation of motion:
vf = vi + at
Where:
vf - final velocity
vi - initial velocity
a - acceleration
t - time
In this case, the car starts from rest, so the initial velocity (vi) is 0 m/s. The deceleration (a) is given as 1.3 m/s^2, and the time (t) is given as 5.6 s.
Substituting the values into the equation, we have:
vf = 0 + (-1.3) * 5.6
Simplifying the equation:
vf = -7.28 m/s
Now, to find the distance traveled by the car during this time interval, we can use another equation of motion:
d = vit + (1/2)at^2
Where:
d - distance
vi - initial velocity
a - acceleration
t - time
Again, substituting the values given:
d = 0 * 5.6 + (1/2) * (-1.3) * (5.6^2)
Simplifying the equation:
d = 0 + (1/2) * (-1.3) * 31.36
d = -20.384 m
It's important to note that the negative sign in the answers indicates that the car is decelerating (slowing down) in the negative direction.