The car's (average) acceleration would be
[tex]a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
The car's position over time would be given by
[tex]x=v_0t+\dfrac12at^2[/tex]
so that after 2.4 seconds, the car will have traveled a distance of
[tex]x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2[/tex]
[tex]\implies x=77.5\,\mathrm m[/tex]
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) = 2.4 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}}{2.4\:s}[/tex]
[tex]\boxed{\boxed{a \approx 11.5\: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) = 11.5 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 * 2.4 + \dfrac{11.5*(2.4)^{2}}{2}[/tex]
[tex]d = 44.4 + \dfrac{11.5*5.76}{2}[/tex]
[tex]d = 44.4 + \dfrac{66.24}{2}[/tex]
[tex]d = 44.4 + 33.12[/tex]
[tex]\boxed{\boxed{d = 77.52\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
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