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Physics .... A particle moves along?
A p0tarticle moves along a path that varies in time according to the following:
x(t)=(10+15.0t^2-25.0t^3)m and y(t)=(15.0+10.t-5.00t^2)m
What is the magnitude of acceleration at t=3.00s
*particle*
2 Answers
- NCSLv 72 years agoFavorite Answer
x(t) = 10 + 15.0t² - 25.0t³
v_x(t) = dx/dt = 30.0t - 75.0t²
a_x(t) = d²x/dt² = 30.0 - 150t
which for t = 3.00 s yields
a_x = -420 m/s²
y(t) = 15.0 + 10t - 5.00t²
v_y(t) = dy/dt = 10 - 10.0t
a_y(t) = d²y/dt² = -10
at all times t
so at t = 3.00s,
|a| = √(420² + 10²) m/s² = 420 m/s²
Hope this helps!
- Anonymous2 years ago
What the heck is a p0tarticle? I've never heard of that.
