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musickatia [10]
2 years ago
8

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 distan

ce traveled.
Physics
2 answers:
solong [7]2 years ago
8 0

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}

________________________________  

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

xeze [42]2 years ago
4 0

a=Δv/Δt=(v-v₀)/t=(46.1-18.5)/2.47=27.6/2.47=11.17 m/s²

d=v₀t+at²/2=18.5*2.47+11.17*2.47²/2=45.69+34.08=79.776 m

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