What is the instantaneous velocity of an object with position at time t equal to # f(t)= ((t-2)^2,5t-3) # at # t=2 #?

Question

What is the instantaneous velocity of an object with position at time t equal to #              f(t)=   ((t-2)^2,5t-3)  # at #  t=2 #?

Aurora Alvarado · Accepted Answer

Instantaneous velocity is #5#.

In a parametric form regarding position of an object #f(t)=(x(t),y(t))#, we get two component of velocity #(dx)/(dt)# and #(dy)/(dt)#, which are at right angles to each other Here we have #f(t)=((t-2)^2,5t-3)# hence two components are #2(t-2)# and #5# and at #t=2# these are #0# and #5# and hence instantaneous velocity is #sqrt(0^2+5^2)=sqrt25=5#.

Evelyn Agard · Answer

undefined To find the instantaneous velocity at $ t = 2 $ for the object with position given by $ f(t) = ((t - 2)^2, 5t - 3) $, we first differentiate the position function to get the velocity function. The velocity function $ v(t) $ is given by:

$$ v(t) = \left( \frac{d}{dt}((t - 2)^2), \frac{d}{dt}(5t - 3) \right) $$

Differentiating each component separately:

$$ v(t) = \left( 2(t - 2) \cdot \frac{d}{dt}(t - 2), 5 \cdot \frac{d}{dt}t \right) $$

$$ v(t) = \left( 2(t - 2) \cdot 1, 5 \cdot 1 \right) $$

$$ v(t) = \left( 2t - 4, 5 \right) $$

Now, to find the instantaneous velocity at $ t = 2 $, substitute $ t = 2 $ into the velocity function:

$$ v(2) = \left( 2(2) - 4, 5 \right) $$

$$ v(2) = \left( 0, 5 \right) $$

So, the instantaneous velocity of the object at $ t = 2 $ is $ (0, 5) $.