A particle p is initially at the point (2, 6) in relation to an origin O, and is moving with velocity (3i + j) m/s. It has a constant acceleration (16i + 24j)m/s^2. Show that after 2 seconds it is moving directly away from O and find its speed at that time

Answer 1

u = x component of speed = 3 + 16 t
v = y component of speed = 1 + 24 t

at t = 2
u = 3 + 32 = 35
v = 1 + 48 = 49

slope of path at t = 2 = v/u = 49/35

location at t = 2:
x = 2 + 3(2) + (1/2)(16)(4) = 40
y = 6 + 1(2) +(1/2)(24)(4) = 56

Tangent to path at t = 2
y = (49/35) x + b
56 = (49/35)40 + b
56 = 56 + b
b = 0
so
y = 1.4 x which goes through the origin

Answer 2

oh, speed at t = 2
we got u = 35 and v = 49
speed = sqrt (35^2 + 49^2)