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

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 = \frac{\Delta{V}}{\Delta{T^}}}`

Solving:

`a = \frac{\Delta{V}}{\Delta{T^}}`

`a = (27.6\:(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 + (a*t^(2))/(2)`

`d = 18.5 * 2.4 + (11.5*(2.4)^(2))/(2)`

`d = 44.4 + (11.5*5.76)/(2)`

`d = 44.4 + (66.24)/(2)`

`d = 44.4 + 33.12`

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

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Answer 2

The car's (average) acceleration would be

`a=(46.1\,(\mathrm m)/(\mathrm s)-18.5\,(\mathrm m)/(\mathrm s))/(2.4\,\mathrm s)=11.5\,(\mathrm m)/(\mathrm s^2)`

The car's position over time would be given by

`x=v_0t+\frac12at^2`

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

`x=\left(18.5\,(\mathrm m)/(\mathrm s)\right)(2.4\,\mathrm s)+\frac12\left(11.5\,(\mathrm m)/(\mathrm s^2)\right)(2.4\,\mathrm s)^2`

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