Instantaneous Velocity

**Purpledog100** · May 11th 2009, 04:11 PM

Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
s(t)= square root of (t+1)
t= 1

This is the only problem I have left and I cannot make much headway in it. Can someone please help me out?

**lmasud** · May 11th 2009, 04:17 PM

Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
s(t)= square root of (t+1)
t= 1

since ur in pre-calc, u probably dont know chain rule, but here is how to solve it:

s(t)= (t+1)^(.5)

v(t)= (.5)((t+1)^(-.5)) , where v(t) represents instantaneous velocity.

v(1)= 1/ 2*(2^.5)

**skeeter** · May 11th 2009, 04:51 PM

Originally Posted by Purpledog100

Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
s(t)= square root of (t+1)
t= 1

This is the only problem I have left and I cannot make much headway in it. Can someone please help me out?

from first principles ...

$v(a) = \lim_{t \to a} \frac{s(t) - s(a)}{t - a}$

$v(1) = \lim_{t \to 1} \frac{s(t) - s(1)}{t - 1}$

$v(1) = \lim_{t \to 1} \frac{\sqrt{t+1} - \sqrt{2}}{t - 1}<br />$

$v(1) = \lim_{t \to 1} \frac{\sqrt{t+1} - \sqrt{2}}{t - 1} \cdot \frac{\sqrt{t+1} + \sqrt{2}}{\sqrt{t+1} + \sqrt{2}}$

$v(1) = \lim_{t \to 1} \frac{(t+1) - 2}{(t - 1)(\sqrt{t+1} + \sqrt{2})}$

$v(1) = \lim_{t \to 1} \frac{t - 1}{(t - 1)(\sqrt{t+1} + \sqrt{2})}<br />$

$v(1) = \lim_{t \to 1} \frac{1}{\sqrt{t+1} + \sqrt{2}} = \frac{1}{2\sqrt{2}}$

**Purpledog100** · May 11th 2009, 06:46 PM

Thank you for the help!

Thread: Instantaneous Velocity

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