A car is traveling at 13 m/s and accelerates at - 0.95 m/s2. What would be the car's velocity after a distance of
9.2 m?

v = 12.3 m / s

Explanation:

This is an exercise in kinetics in one dimension

        v² = v₀² + 2 a x

In this exercise they tell us that the initial velocity is (v₀ = 13 m / s), the acceleration is a = -0.95 m / s2 and the distance x = 9.2 m

we substitute

         v = √ (13 2 - 2 0.95 9.2)

         v = 12.3 m / s

note that as the acceleration is negative the vehicle is stopping

spectral lines lyman is one set

truck y has a greater velocity than truck x.