How do i find the tangential acceleration??

Question

If the motion of a particle is defined by:
x = 7t^2 + 6t +9 and
y = 6t^3 -9
where x and y are in m, and t is in secs

what is the component of acclereation in the tangetial direction when t=2.0s??

From what i understand,
dx/dt= 14 and dy/dt= 36t. If i substitute t= 2.0 in both these equations, and find the magnitude of this, then a= 73.34m/s^2... How do i find the acceleration in the TANGETIAL direction though..

Help please,
many thanks

Answer 1

You get the component of a vector into the direction of any other by projecting it on the other vector. Mathematically this is done by building the scalar product with the unit vector into the asked direction. Now you just need to know that the velocity is tangential in each point of the path. So calculate the vectorial velocity, divide it by it's absolute value to get the unit vector and multiply that with the vector of the acceleration.

Answer 2

x = 7t^2 + 6t +9
y = 6t^3 -9

dx/dt = 14t + 6
dy/dt = 18t^2

d2x/dt2 = 14
d2y/dt2 = 36t

So, at time t=2, the velocity is v = (34, 72) and the acceleration is a = (14, 72).
The component of acceleration in the direction of v (which is the tangential direction) is given by
(a.v) v / ||v||^2
where (a.v) represents the vector dot product: so it is
[(34×14 + 72×72) / (34^2 + 72^2)] (34, 72)
= 5660/6340 (34, 72)
= (30.4, 64.3) to 1 d.p.

You can also convert v into a unit vector u = v / ||v|| = (0.4270, 0.9042). Then the acceleration in the tangential direction is simply (a.u) u = (71.08) (0.4270, 0.9042) = (30.4, 64.3) as before.

If you only want the magnitude of the acceleration in the tangential direction, it is (a.v) / ||v|| = 71.08 m/s^2.