Question

A point has a position vector given by r of t equals t^2+1, 4 minus t, t for all time t ≥ 0 seconds. Find the speed of the object at t = 1 seconds. (10 points)

A) 1
B) 5
C) 2
D) square root of 5

Answer 1

do you mean x = t^2 + 1

and y = 4 - t
so
R = (t^2+1) i + (4-t) j
????
if so
dR/dt = 2 t i - 1 j
speed = sqrt (Vx^2+Vy^2) = sqrt [ 4 t^2 +1*1 ]
at t = 1
speed = sqrt (5)