Posted by **Jat** on Sunday, February 26, 2012 at 3:39am.

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

- Please help me with M1 kinematics! -
**Damon**, Sunday, February 26, 2012 at 4:16am
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

- Please help me with M1 kinematics! -
**Damon**, Sunday, February 26, 2012 at 4:39am
oh, speed at t = 2

we got u = 35 and v = 49

speed = sqrt (35^2 + 49^2)

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