Question

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.

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) = 2.47 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))/(2.47\:s)`

`\boxed{\boxed{a \approx 11.174\: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) = 11.174 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 * 2.47 + (11.174*(2.47)^(2))/(2)`

`d = 45.695 + (11.174*6.1009)/(2)`

`d = 45.695 + (68.1714566)/(2)`

`d = 45.695 + 34.0857283`

`d = 79.7807283 \to \boxed{\boxed{d \approx 79.8\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\green{\checkmark}`

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