The position vector r of a particle moving in an xy plane is given by vector r =

Question

The position vector r of a particle moving in an xy plane is given by vector r = (4.00tt^3 - 6.00t)cap i + (7.00 - 8.00t^4)cap j, with vector r in meters and t in seconds. In unit-vector notation, calculate the following for t = 3.00 s. (a) vector r = m (b) vector v = m/s (c) vector a = m/s^2 (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3.00 s? (counterclockwise from the +x-axis) degree A particle moves so that its position (in meters) as a function of time (in seconds) is vector r = cap i + 3t^2 cap j + 5t cap k. (Use the following as necessary: t.) (a) Write an expression for its velocity as a function of time. vector v(t) = m/s (b) Write an expression for its acceleration as a function of time. vector a = m/s^2

Explanation / Answer

5. r = (4 t^3 - 6t)i + (7 - 8 t^4)j

v = dr/dt = (12 t^2 - 6)i + ( - 32 t^3) j

a = dv/dt = (12t)i + (-96 t^2)j

at t = 3 s

(A) r = 90i - 641j m

v = 102i - 864j m/s

a = 36i - 864j m/s^2

(B) slope = dy/dx

y = 7 - 8 t^4 and x = 4 t^3 - 6t

dy/dx = - 864 / 102 = - 8.47

8.47 = tan^-1(theta)

theta = 83.3 deg

from +v x axis CCw = 360 - 83.3 = 276.7 deg

6. v = dr/dt = 6t j + 5 k

a = dv/dt = 6 j

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