A car slowed down at a constant rate from 32meters per second to 3.0 meters per second in 9.0 seconds. What was the distance the car traveled in 9.00 seconds?
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You only need two steps for this.
Multiply 9.0 seconds by the AVERAGE velocity during the interval, which is (32 + 3)/2 = 17.5 m/s in this case.
9.0 * 17.5 = ____ m
v(final)=v,
v(initial) =vₒ,
v = vₒ -a•t,
a= (vₒ- v)/t=(32-3)/9=29/9=3.22 m/s².
s=vₒ•t- a•t ²/2=
=32•9 - 3.22•81/2 =157.6 m
Oh my goodness, thank you so much! I was trying to do so much more then I needed too!
To find the distance the car traveled in 9.0 seconds, we can use the formula for average velocity:
Average Velocity = (Initial Velocity + Final Velocity) / 2
In this case, the initial velocity (u) is 32 meters per second and the final velocity (v) is 3.0 meters per second. Substituting these values into the formula, we get:
Average Velocity = (32 + 3.0) / 2
Average Velocity = 35.0 / 2
Average Velocity = 17.5 meters per second
Now, we know that average velocity is equal to total displacement (distance traveled) divided by the time interval. Rearranging this formula, we can solve for distance:
Distance = Average Velocity * Time
Substituting the values of average velocity (17.5 meters per second) and time (9.0 seconds) into the formula, we find:
Distance = 17.5 * 9.0
Distance = 157.5 meters
Therefore, the car traveled a distance of 157.5 meters in 9.0 seconds.