the velosity of a particle moving along the x axis varies according to the expression Vx=40-5t2 whereVx is metre per second and t is in seconds the average accelaration in the time interval t=0 to t=2s is​

9514 1404 393

  -10 m/s²

Explanation:

The acceleration is the rate of change of velocity. The average rate of change of velocity will be the difference over the interval, divided by the width of the interval.

 (V(2) -V(0))/(2 -0) = (40 -5(2²)) -(40 -5(0²)) = (20 -40)/2 = -10 . . . m/s²

The average acceleration on [0, 2] is -10 m/s².

Additional comment

The average rate of change of a quadratic function on an interval is exactly equal to the derivative of the function at the midpoint of the interval:

  Vx'(t) = -10t   ⇒  Vx'(1) = -10 . . . m/s² (midpoint of [0, 2] is t=1)

d is the correct answer