Respuesta :
A [tex]1000[/tex] kg car travelling with velocity [tex]70[/tex]m/s takes [tex]3[/tex] to stop under full braking. And the same car under similar road conditions travelling with velocity [tex]140m/s[/tex] takes [tex]12m[/tex] to stop under full braking.
How can we find Distance at velocity [tex]140m/s[/tex] ?
In case of [tex]70m/s[/tex] car stop under full braking after [tex]3m[/tex].
So we can find the acceleration by third equation of motion
[tex]v^2=u^2+2as[/tex]
Here [tex]v[/tex] is final velocity which is zero.
[tex]u[/tex] is initial velocity which is [tex]70m/s[/tex]
[tex]s[/tex] is displacement
So substitute the values in equation of motion
[tex]0=70^2+2*a*3\\-4900/6=a\\-816.6m/s^2=a[/tex]
In case of initial velocity [tex]140m/s[/tex] acceleration is same because same car and same road condition so acceleration is [tex]-816.6m/s^2[/tex]
We can find the distance by using the equation of motion
[tex]v^2=u^2+2as\\0=140^2+2*(-816.6)*s\\-19600/-1633.2=s\\12m=s[/tex]
Distance is [tex]12m[/tex]
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