Question

A particle moves along the x-axis, with x given in meters. Its position is described by the following equation: x(t) = 6 + 4t - 7t^2.

A) Determine its position, velocity, and acceleration at t = 3s
a) Position: 6 meters, Velocity: 18 m/s, Acceleration: -14 m/s^2
b) Position: 3 meters, Velocity: 12 m/s, Acceleration: -14 m/s^2
c) Position: 6 meters, Velocity: 18 m/s, Acceleration: 14 m/s^2
d) Position: 3 meters, Velocity: 12 m/s, Acceleration: 14 m/s^2

Answer 1

At t = 3s, the position of the particle is -45 meters, the velocity is -38 m/s, and the acceleration is -14 m/s^2.

Step-by-step explanation:

To determine the position, velocity, and acceleration of the particle at t = 3s, we need to substitute t = 3 into the given equation x(t) = 6 + 4t - 7t^2.

1. Position: Plugging in t = 3, we get x(3) = 6 + 4(3) - 7(3)^2 = 6 + 12 - 63 = -45 meters.
2. Velocity: Velocity is the derivative of position with respect to time. Taking the derivative of x(t) = 6 + 4t - 7t^2, we get v(t) = 4 - 14t. Plugging in t = 3, we get v(3) = 4 - 14(3) = -38 m/s.
3. Acceleration: Acceleration is the derivative of velocity with respect to time. Taking the derivative of v(t) = 4 - 14t, we get a(t) = -14. The acceleration is constant at -14 m/s^2.