Respuesta :
from kinematics equation if we know that final speed is ZERO and initial speed is given that due to constant deceleration the object will stop in some distance "d" and this distance can be calculated by kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - v^2 = 2(a)d[/tex]
here acceleration due to friction will be same at all different speed
so for 45 km/h speed the distance of stop is 15 m
while at other speed 112.5 km/h the distance will be unknown
now we will have
[tex]0 - 45^2 = 2(a)15[/tex]
[tex]0 - 112.5^2 = 2(a)d[/tex]
now divide above two equations
[tex]\frac{45^2}{112.5^2} = \frac{15}{d}[/tex]
[tex]d = 93.75 m[/tex]
So it will stop in distance 93.75 m
Answer:
The distance travelled through skiding by car is [tex]93.75\;\rm{m[/tex]
Explanation:
Given: A car moving at [tex]45\;\rm{km/h}[/tex] skids [tex]15\;\rm{m[/tex] with locked brakes.
As per question,
Final speed [tex](v)[/tex] is [tex]0\;\rm{m/s[/tex] .
Intial speed [tex](u)[/tex] is [tex]45\;\rm{km/hr}=45\times\frac{5}{18}=12.5\;\rm{m/s[/tex]
And speed with locked brakes: [tex]112.5\;\rm{km/hr}=112.5\times\frac{5}{18}=31.25\;\rm{m/s[/tex]
Using the formula of laws of motion:
[tex]v^2-u^2=2as[/tex]
Substiuting the values we get:
[tex]0-(12.5)^2=2\times{a}\times15\;\;\;\;\;.....(1)\\0-(31.25)^2=2\times{a}\times{d\;\;\;\;\;.....(2)\\[/tex]
Dividing the equation (1) and (2):
[tex]\frac{156.25}{976.5625}=\frac{15}{d}\\[/tex]
[tex]d=93.75\;\rm{m[/tex]
Hence, distance travelled through skiding by car is [tex]93.75\;\rm{m[/tex].
Learn more about laws of motion here:
https://brainly.com/question/15754094?referrer=searchResults
