Question

A car travelling at 22.4m/s skids to a stop in 2.55s. Determine the skidding distance of the car.

Answer 1

Hello!

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car

We have the following data:

a (acceleration) = ? (in m/s²)

t (time) = 2.55 s

Vf (final velocity) = 22.4 m/s

Vi (initial velocity) = 0 m/s

We apply the data to the formula of the hourly function of the velocity, let's see:

$V_f = V_i + a*t$

$22.4 = 0 + a*2.55$

$22.4 = 2.55\:a$

$2.55\:a = 22.4$

$a = \dfrac{22.4}{2.55}$

$\boxed{a \approx 8.784\:m/s^2}\Longrightarrow(acceleration)$

*** The distance traveled ?  

We have the following data:  

Vf (final velocity) = 22.4 m/s

Vi (initial velocity) = 0 m/s  

a (average acceleration) = 8.784 m/s²  

d (distance interval) = ? (in m)

By the formula of the space of the Uniformly Varied Movement, it is:

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

$d = 0 * 2.55 + \dfrac{9.955*(2.55)^{2}}{2}$

$d = 0 + \dfrac{8.784*6.5025}{2}$

$d = \dfrac{57.11796}{2}$

$d = 28.55898 \to \boxed{\boxed{d \approx 28.56\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}$

*** Another way to solve:

Since we do not need to know the time elapsed during the movement, we apply the data of the question to the Equation of Torricelli, let us see:

$V_f^2 = V_i^2 + 2*a*d$

$22.4^2 = 0^2 + 2*8.784*d$

$501.76 = 17.568\:d$

$17.568\:d = 501.76$

$d = \dfrac{501.76}{17.568}$

$\boxed{\boxed{d \approx 28.56\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}$

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