Question

A particle moves along the curve x= 3t²- 2 , y = 2t -t^2, z=t^3+4 Find the
velocity and acceleration at t=4.​

Answer 1

Answer:

VELOCITY = 48 m/ s

ACCELERATION = 24.82 m/ s^2.

Step-by-step explanation:

$\frac{dx}{dt} = 6t = 6 \times 4 = 24 \\ \\ \frac{dy}{dt} = 2 - 2t = 2 - 2 \times 4 = 2 - 8 = - 6 \\ \\ \frac{dz}{dt} = 3 {t}^{2} = 3 \times 16 = 48 \\ \\ \\ v = \sqrt{ {24}^{2} + {6}^{2} + {48}^{2} } = \sqrt{576 + 36 + 2304} = \sqrt{2304} = 48 \: m {s}^{ - 1} \\ \\ \frac{ {d}^{2}x }{d {t}^{2} } = 6 \\ \\ \frac{ {d}^{2} y}{d {t}^{2} } = - 2 \\ \\ \frac{ {d}^{2}z }{d {t}^{2} } = 6t = 6 \times 4 = 24 \\ \\ \\ acceleration = \sqrt{ {6}^{2} + {2}^{2} + {24}^{2} } = \sqrt{36 + 4 + 576} = 24.82 \: m {s}^{ - 2} .$