Question

A particle moves along a curve whose parametric equations are x = t³ + 2t, y = -3e^-2t and z = 2 sin (5t), where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s²) at t = 0 is ____.

Answer 1

x = t³ + 2t

\(\therefore {V_x} = \frac{{dx}}{{dt}} = 3{t^2} + 2\)

\({a_x} = \frac{{{d^2}x}}{{d{t^2}}} = 6t\)

y = -3e^-2t

v_y = +6e^-2t

\({a_y} = \frac{{{d^2}s}}{{d{t^2}}} = - 12{e^{ - 2t}}\)

Z = 2 sin (5t)

∴ V_z = 10 cos 5t

a_z = -50 sin 5t

at, t = 0

a_x = 0

a_y = -12

a_z = 0

\(\therefore Acceleration = \sqrt {a_x^2 + a_y^2 + a_z^2} \)

\(= \sqrt {0 + {{\left( { - 12} \right)}^2} + 0} = 12\)