Answer:
(a) the particle position = 135 m
(b) the velocity of the particle = 44 m/s
(c) the acceleration of the particle = 50 m/s²
Explanation:
Solution to Question 2.
Given;
velocity of a particle, v = 2 - 4t + 2t³
initial position at t = 0, s₀ = 3 m
(a) the particle position at t = 3 s
s = vt
s = (2 - 4t + 2t³)t
s = 2t - 4t² + 2t⁴
s = s₀ + s₃
s = s₀ + 2(3) - 4(3²) + 2(3⁴)
s = s₀ + 6 - 36 + 162
s = s₀ + 132 m
s = 3m + 132 m
s = 135 m
(b) the velocity of the particle at t = 3 s
v = 2 - 4t + 2t³ = 2 - 4(3) + 2(3)³
v = 44 m/s
(c) the acceleration of the particle at t = 3s
v = 2 - 4t + 2t³
[tex]a = \frac{dv}{dt} \\\\a = \frac{d}{dt} (2 -4t + 2t^3)\\\\a = -4 + 6t^2[/tex]
a = -4 + 6(3)²
a = 50 m/s²
