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:  

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

Solving:

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

$a = \dfrac{27.6\:\dfrac{m}{s}}{2.47\:s}$

$\boxed{\boxed{a \approx 11.174\:m/s^2}}\longleftarrow(acceleration)\:\:\:\:\:\:\bf\green{\checkmark}$

* 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:

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

$d = 18.5 * 2.47 + \dfrac{11.174*(2.47)^{2}}{2}$

$d = 45.695 + \dfrac{11.174*6.1009}{2}$

$d = 45.695 + \dfrac{68.1714566}{2}$

$d = 45.695 + 34.0857283$

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

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Answer 2

$\huge{ \bold{ \underline{ \red{ \: hey \: mate}}}} : \: )$

HERE IS YOUR ANSWER BELOW......,

$\huge{ \: u=18.5 m/s}$
$\huge{ \: v=46.1 m/s}$
$\huge{ \: t=2.47 secs}$

${ \large{ \: a=(v-u)/t}}......................formula$

$\large{ \: = > (46.1-18.5)/ 2.47}$

$\large{ \: = > 27.6/2.47}$

$\large{ \: = > 11.17m/s^2}$

So,
$\small{ \bold{ \underline{ \blue{ \: the \: accelaration \: is 11.17 m/s^2}}}}$






Now The Distance..,

as we know...,,



Distance= Vi × t



d = 18.5 × 2.47 +
11.174 × (2.47) / 2



d = 45.695 +
11.174 × 6.1009/2


d = 45.695 +
68.1714566/2



d = 45.695 + 34.0857283



d = 79.7807283

i.e
d = 79.8m




Hence the distance traveled is 79.8 metres.







#BeBrainly....................!










$\small{ \bold{ \underline{ \red{ \: plz \: mark \: as \: the \: brainliest}}}}$








HOPES HELP...............








ByIshita..,
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