A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is
v(t)=1t^1/2+3
.
Find the car's average velocity (in m/s) between t=5 and t=8.
If you could help me out with the steps of this problem I would greatly appreciate it!
2007-04-25 17:55:38 · 4 answers · asked by Mark S 1 in Science & Mathematics ➔ Mathematics
v(5) = 5^(1/2) + 3 = 5.236 m/s
v(8) = 8^(1/2) + 3 = 5.828 m/s
v = (v(5) + v(8))/2 = 5.532 m/s
2007-04-25 18:05:40 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The average velocity can between t=5 and t=8 can be found by evaluating (v(8) - v(5))/(8-5).
v(5) = 5^(1/2) + 3
v(8) = 8^(1/2) + 3
So, v(8) - v(5) = (8^(1/2) + 3) - (5^(1/2) + 3)
= 8^(1/2) - 5^(1/2)
Therefore the average velocity between t=5 and t=8 is
[8^(1/2) - 5^(1/2)]/3.
2007-04-26 01:03:08 · answer #2 · answered by polymac98 2 · 0⤊ 1⤋
v = 3 + t^(1/2)
8
{â«[3 + t^(1/2)]dt}/3 =
5
[3*8 - 3*5 + (2/3)(8^(3/2) - 5^(3/2)]/3 =
[9 + (2/3)(11.44708)]/3 =
v(avg) = 5.543795 m/s
2007-04-26 01:42:47 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
hey. are you trying to ask someone to do your homework ?
2007-04-26 01:11:22 · answer #4 · answered by modern wushu 2 · 0⤊ 0⤋