Question

For each value below, enter the number correct to four decimal places. Suppose an arrow is shot upward on the moon with a velocity of 52 m/s, then its height in meters after t t seconds is given by h ( t ) = 52 t − 0.83 t 2 h(t)=52t-0.83t2. Find the average velocity over the given time intervals.

Answer 1

The time intervals are missing in the question.You can put any time in equation v(t) which I have solved

Answer:

`v(t)=52-1.66t`

Step-by-step explanation:

Given data

Velocity=52 m/s

Height h(t)=52t-0.83t²

To find

Average velocity

Solution

As we know that velocity is first derivative of distance with respect to time

So

`v(t)=(dh(t))/(dt)\\ v(t)=(d)/(dt)[52t-0.83t^(2) ]\\ v(t)=52-1.66t`

After one second the velocity is

`v(t)=52-1.66t\\at\\t=1seconds\\v(1)=52-1.66(1)\\v(1)=56.33m/s`