Hello!
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
* Determine the acceleration of the car....
We have the following data:
V (final velocity) = 46.1 m/s
Vo (initial velocity) = 18.5 m/s
ΔV (speed interval) = V - Vo → ΔV = 46.1 - 18.5 → ΔV = 27.6 m/s
ΔT (time interval) = 2.47 s
a (average acceleration) = ? (in m/s²)
Formula:
[tex]\boxed{a = \dfrac{\Delta{V}}{\Delta{T^}}}[/tex]
Solving:
[tex]a = \dfrac{\Delta{V}}{\Delta{T^}}[/tex]
[tex]a = \dfrac{27.6\:\dfrac{m}{s}}{2.47\:s}[/tex]
[tex]\boxed{\boxed{a \approx 11.174\:m/s^2}}\longleftarrow(acceleration)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
* The distance traveled ?
We have the following data:
Vi (initial velocity) = 18.5 m/s
t (time) = 2.47 s
a (average acceleration) = 11.174 m/s²
d (distance interval) = ? (in m)
By the formula of the space of the Uniformly Varied Movement, it is:
[tex]d = v_i * t + \dfrac{a*t^{2}}{2}[/tex]
[tex]d = 18.5 * 2.47 + \dfrac{11.174*(2.47)^{2}}{2}[/tex]
[tex]d = 45.695 + \dfrac{11.174*6.1009}{2}[/tex]
[tex]d = 45.695 + \dfrac{68.1714566}{2}[/tex]
[tex]d = 45.695 + 34.0857283[/tex]
[tex]d = 79.7807283 \to \boxed{\boxed{d \approx 79.8\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
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