# A truck is moving along at 80 km/h when it hits a gravel patch which causes it to accelerate at -5.0 km/s. How far will the truck travel before it slows to 20.0 km/h?

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For an object traveling at an initial velocity u and which is accelerating at a rate of acceleration a, the velocity after traveling a distance s is v. The different quantities are related as: `v^2 - u^2 = 2*a*s`

Substituting the values given, we get:

`20^2 - 80^2 = 2*(-5*3600)*s`

=> s = 400 - 6400 = -36000*s

=> s = `-6000/-36000`

=> s = `1/6` km

**The truck reaches a speed of 20 km/h after traveling for 1/6 km on the gravel patch.**

Initial velocity (u)=80 km/h

Final velocity(V)=20 km/h

Acceleration(a)= -5 km/s= -5*3600 km/h

Distance(s)=?

using third equation of motion:

v2 – u2 = 2.a.s

202 - 802= 2 (-5*3600) s

=> 400-6400 = -36000.s

=> -6000= -36000.s

=> s= -6000 / -36000

s=1/6 km

s= 166.666666667 m

Therefore the truck will travel 1/6 km or 166.666666667 metres till reaching a speed of 20 km/h after hitting gravel patch.