Answer:
a) 2 m/s
b) i) [tex]K.E = 50 (1.5t^2 + 2) ^2\\[/tex]
ii) [tex]F = 3tm[/tex]
Explanation:
The function for distance is [tex]x = 0.5t ^3 + 2t[/tex]
We know that:
Velocity = [tex]v= \frac{d}{dt} x[/tex]
Acceleration = [tex]a= \frac{d}{dt}v[/tex]
To find speed at time t = 0, we derivate the distance function:
[tex]x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2[/tex]
Substitute t = 0 in velocity function:
[tex]v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2[/tex]
Velocity at t = 0 will be 2 m/s.
To find the function for Kinetic Energy of the box at any time, t.
[tex]Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\[/tex]
We know that [tex]Force = mass \times acceleration[/tex]
[tex]a = v'(t) = 1.5t^2 + 2\\a = 3t[/tex]
[tex]F = m \times a\\F= m \times 3t\\F = 3tm[/tex]
