A point has a position vector given by r(t)= (t^2 +1,4-t), t for all time t ≥ 0 seconds. Find the speed of the object at t = 1 seconds
a) 1
b) 5
c) 2
d) √5
To find the speed of an object at a specific time, we need to find the magnitude of its velocity vector at that time. The velocity vector is the derivative of the position vector with respect to time.
Given that the position vector is given by r(t) = (t^2 + 1, 4 - t), we can find the velocity vector v(t) by taking the derivative of each component of the position vector:
v(t) = (d/dt (t^2 + 1), d/dt (4 - t))
Differentiating each component with respect to t:
v(t) = (2t, -1)
Now, to find the speed at t = 1 second, substitute t = 1 into the velocity vector:
v(1) = (2(1), -1)
= (2, -1)
The magnitude of this velocity vector is given by the Pythagorean theorem:
|v(1)| = sqrt((2^2) + (-1^2))
= sqrt(4 + 1)
= sqrt(5)
Therefore, the speed of the object at t = 1 second is √5, which corresponds to option d).
this is just like your previous post, only easier.
Whatcha got?