A particle travels along a straight line with a velocity 2 v t m s   (12 3) /, where t is in seconds. When t s  1, the particle is located 10m to the left of the origin. Determine the acceleration when t s  4, the displacement from t s  0 to t s  10, and the distance the particle travels during this time period. 2 a m s   24 /, s m   880, d m

a) a = - 24 m / s², b) Δx = 1180 m

Explanation:

a) This is a one-dimensional kinematics exercise, let's use the equation

x = x₀ + v₀ t + ½ a t²

in this case give us the initial velocity v₀ = 2 m / s and the location (x = - 10m) for t = 1 s, find us the acceleration

a = (x-x₀ - v₀ t) 2 / t²

let's calculate

a = (-10 - 0 - 2 1) 2/1²

a = - 24 m / s²

The sign of acceleration indicates that it is heading to the left; this acceleration is constant throughout the movement

therefore the acceleration for t = 4 s is also a = 24 m / s²

b) The displacement between t = 0 and t = 10 s

for t = 0 it is at x = xo

for t = 10 x = xo + 2 10 - ½ 24 10²

x = xo + 20 - 1200

x (10) = xo - 1180

in displacement is

Δx = x (0) - x (10)

Δx = 1180 m