A particle moves on a line away from its initial position so that after t hours it is s = 6t^2 + 2t miles from its initial position. Find the average velocity of the particle over the interval [1, 4]. Include units in your answer.
after 1 sec, d = 6(1) + 2 = 8
after 4 sec, d = 6(16) + 2(4) = 104
avg velocity = total distance/total time
= (104-8)(4-1) = 32 miles/hr
To find the average velocity of the particle over the interval [1, 4], we need to calculate the change in position and divide it by the change in time.
First, let's find the position of the particle at t = 1 and t = 4 using the given equation s = 6t^2 + 2t.
When t = 1:
s(1) = 6(1)^2 + 2(1)
s(1) = 6 + 2
s(1) = 8 miles
When t = 4:
s(4) = 6(4)^2 + 2(4)
s(4) = 6(16) + 8
s(4) = 96 + 8
s(4) = 104 miles
Next, we calculate the change in position:
Change in position = s(4) - s(1)
Change in position = 104 - 8
Change in position = 96 miles
Now, let's calculate the change in time:
Change in time = 4 - 1
Change in time = 3 hours
Finally, we divide the change in position by the change in time to find the average velocity:
Average velocity = Change in position / Change in time
Average velocity = 96 miles / 3 hours
Average velocity = 32 miles/hour
Therefore, the average velocity of the particle over the interval [1, 4] is 32 miles/hour.