Respuesta :
a = ( V2 - V1)/( t2 - t1)
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
Hello!
How long will it take a car to accelerate from 15.2 m/s to 23.5 m/s if the car has an average acceleration of 3.2 m/s² ?
We have the following data:
Vf (final velocity) = 23.5 m/s
Vi (initial velocity) = 15.2 m/s
ΔV (speed interval) = Vf - Vi → ΔV = 23.5 - 15.2 → ΔV = 8.3 m/s
ΔT (time interval) = ? (in s)
a (average acceleration) = 3.2 m/s²
Formula:
[tex]a = \dfrac{\Delta{V}}{\Delta{T^}}[/tex]
Solving:
[tex]a = \dfrac{\Delta{V}}{\Delta{T^}}[/tex]
[tex]3.2 = \dfrac{8.3}{\Delta{T^}}[/tex]
[tex]\Delta{T^} = \dfrac{8.3}{3.2}[/tex]
[tex]\Delta{T^} = 2.59375 \to \boxed{\boxed{\Delta{T^} \approx 2.6\:s}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
Answer:
The car will take approximately 2.6 seconds to accelerate
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I Hope this helps, greetings ... Dexteright02! =)