Kinematics Equation 3

Question

A car which is traveling at a velocity of 17 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 250 m. How fast is it going after that acceleration?

Answer 1

Given:

Vo = 17 m/s (Initial Velocity)

a = 5.5 m/s^2 (Acceleration)

Xo = 0

X = 250 m

Find:

V = ?

Solution:

2*a*(X-Xo)+Vo^2 = V^2

2 * 5.5 * (250-0) + 17^2 = V^2

V^2 = 3039

Solution:

V = 55.127 m/s

Answer 2

Use the following kinematic equation

v²=v₀²+2aΔx

where,

a = 5.5 m/s^2,

v₀ = 17 m/s

Δx= 250 m

You'll need to calculate 'v'. Plug in the numbers into the equation to get

v²=17²+2(5.5)(250)

v²=3039

v=55.1 m/s

So the car travels at 55.1 m/s after the acceleration.