Question

A car traveling at 22.4 m/s skids to a stop in 2.25s. Determine the skidding distance of the car

Answer 1

Hello!

A car traveling at 22.4 m/s skids to a stop in 2.25 s. Determine the skidding distance of the car

We have the following data:

a (acceleration) = ? (in m/s²)

t (time) = 2.25 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:

$V_f = V_i + a*t$

$22.4 = 0 + a*2.25$

$22.4 = 2.25\:a$

$2.25\:a = 22.4$

$a = \dfrac{22.4}{2.25}$

$\boxed{a \approx 9.955\:m/s^2}\Longrightarrow(acceleration)$

*** The distance traveled ?  

We have the following data:  

Vi (initial velocity) = 0 m/s  

t (time) = 2.25 s  

a (average acceleration) = 9.955 m/s²  

d (distance interval) = ? (in m)

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

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

$d = 0 * 2.25 + \dfrac{9.955*(2.25)^{2}}{2}$

$d = 0 + \dfrac{9.955*5.0625}{2}$

$d \approx \dfrac{50.4}{2}$

$\boxed{\boxed{d \approx 25.2\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}$

*** Another way to solve:

We have the following data:  

Vf (final velocity) = 22.4 m/s

Vi (initial velocity) = 0 m/s  

a (average acceleration) = 9.955 m/s²  

d (distance interval) = ? (in m)

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:

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

$22.4^2 = 0^2 + 2*9.955*d$

$501.76 = 19.91\:d$

$19.91\:d = 501.76$

$d = \dfrac{501.76}{19.91}$

$\boxed{\boxed{d \approx 25.2\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}$

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

Answer 2

$\bold{Hello}$


Given :

u = 22.4 m/s

v = 0

t = 2.25

s = ?



applying the equation of motion :


$\bold{v = u + at}$


putting the values , we get value of
$\bold{a = 9.955 m/{s}^2}$


putting value of a in other equation . :-


=> $\bold{v^2 = u^2 + 2as}$


by solving this , we get stopping distance :-


$\bold{\large{s = 25.201 m}}$






Hope It Helps You. ^_^