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

magnified images possible images can be real = projected onto a screen

i think it's 35 meters