Respuesta :
[tex]\bold{Hello}[/tex]
Given :
u = 22.4 m/s
v = 0
t = 2.25
s = ?
applying the equation of motion :
[tex]\bold{v = u + at}[/tex]
putting the values , we get value of
[tex]\bold{a = 9.955 m/{s}^2}[/tex]
putting value of a in other equation . :-
=> [tex]\bold{v^2 = u^2 + 2as}[/tex]
by solving this , we get stopping distance :-
[tex]\bold{\large{s = 25.201 m}}[/tex]
Hope It Helps You. ^_^
Given :
u = 22.4 m/s
v = 0
t = 2.25
s = ?
applying the equation of motion :
[tex]\bold{v = u + at}[/tex]
putting the values , we get value of
[tex]\bold{a = 9.955 m/{s}^2}[/tex]
putting value of a in other equation . :-
=> [tex]\bold{v^2 = u^2 + 2as}[/tex]
by solving this , we get stopping distance :-
[tex]\bold{\large{s = 25.201 m}}[/tex]
Hope It Helps You. ^_^
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:
[tex]V_f = V_i + a*t[/tex]
[tex]22.4 = 0 + a*2.25[/tex]
[tex]22.4 = 2.25\:a[/tex]
[tex]2.25\:a = 22.4[/tex]
[tex]a = \dfrac{22.4}{2.25}[/tex]
[tex]\boxed{a \approx 9.955\:m/s^2}\Longrightarrow(acceleration)[/tex]
*** 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:
[tex]d = v_i * t + \dfrac{a*t^{2}}{2}[/tex]
[tex]d = 0 * 2.25 + \dfrac{9.955*(2.25)^{2}}{2}[/tex]
[tex]d = 0 + \dfrac{9.955*5.0625}{2}[/tex]
[tex]d \approx \dfrac{50.4}{2}[/tex]
[tex]\boxed{\boxed{d \approx 25.2\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]
*** 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:
[tex]V_f^2 = V_i^2 + 2*a*d[/tex]
[tex]22.4^2 = 0^2 + 2*9.955*d[/tex]
[tex]501.76 = 19.91\:d[/tex]
[tex]19.91\:d = 501.76[/tex]
[tex]d = \dfrac{501.76}{19.91}[/tex]
[tex]\boxed{\boxed{d \approx 25.2\:m}}\longleftarrow(distance)\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]
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[tex]\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}[/tex]