Answer:
So the car would skid 4 times the initial skidding displacement under the identical conditions when the car's initial velocity is just double.
Explanation:
Under identical conditions when the brakes are applied for the same body moving with twice the initial speed then we can analyse using the equation of motion.
Initially:
[tex]v^2=u^2-2a.s[/tex]
[tex]\Rightarrow s=\frac{u^2}{2a}[/tex] ...........................(1)
where:
[tex]v=[/tex] final velocity of the car = 0 (because the car stops)
[tex]u=[/tex] initial velocity of the car
[tex]a=[/tex] acceleration of the car (here it is deceleration and hence taken with negative sign)
[tex]s=[/tex] displacement of the car before stopping
Final condition:
Initial velocity is, [tex]2u[/tex]
So,
[tex]v^2=(2u)^2-2a.s'[/tex]
[tex]0^2=4u^2-2a.s'[/tex]
[tex]s'=\frac{4u^2}{2a}[/tex]
[tex]s'=\frac{2u^2}{a}[/tex] .............................(2)
Now divide eq. (2) by (1)
[tex]\frac{s'}{s}=4[/tex]
So the car would skid 4 times the initial skidding displacement under the identical conditions when the car's initial velocity is just double.
