Question

a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance traveled

Answer 1

The car's (average) acceleration would be

$a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}$

The car's position over time would be given by

$x=v_0t+\dfrac12at^2$

so that after 2.4 seconds, the car will have traveled a distance of

$x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2$

$\implies x=77.5\,\mathrm m$

Answer 2

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:

$\boxed{a = \dfrac{\Delta{V}}{\Delta{T^}}}$

Solving:  

$a = \dfrac{\Delta{V}}{\Delta{T^}}$

$a = \dfrac{27.6\:\dfrac{m}{s}}{2.4\:s}$

$\boxed{\boxed{a \approx 11.5\:m/s^2}}\longleftarrow(acceleration)\:\:\:\:\:\:\bf\green{\checkmark}$

* 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:

$d = v_i * t + \dfrac{a*t^{2}}{2}$

$d = 18.5 * 2.4 + \dfrac{11.5*(2.4)^{2}}{2}$

$d = 44.4 + \dfrac{11.5*5.76}{2}$

$d = 44.4 + \dfrac{66.24}{2}$

$d = 44.4 + 33.12$

$\boxed{\boxed{d = 77.52\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\green{\checkmark}$

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