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 travelled

Answer 1

v=u+at

46.1=18.5+2.47a

27.6=2.47a

11.17m/s^2 =~a

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) = 2.47 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:

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

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

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

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

$d = 45.695 + 34.0857283$

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

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