Answer to Question #265175 in Physics for Boonj98

Question

A particle moves along the x axis according to

the equation x = 2.00 + 3.00 t - 1.00

, where x is in

meters and t is in seconds. At t = 3.00 s, find (a) the

position of the particle, (b) its velocity, and (c) its acceleration.

Answer 1

I assume, the equation is the following:

$$x(t) = 2+3t-t^2$$

The velocity is the first derivative of the position with respect to time:

$$v(t) =x'(t) = 3-2t$$

The acceleration is the second derivative of the position with respect to time:

$$a(t) = x''(t) = v'(t) =- 2$$

a) Position at $t = 3s$:

$$x(3) = 2 + 3\cdot 3-3^2= 2.00m$$

b) Velocity at $t = 3s$:

$$v(3) = 3-2\cdot 3 = -3.00m/s$$

c) Acceleration at $t = 3s$:

$$a(3) = -2.00m/s^2$$

Answer a) 2.00m, b) -3.00m/s, c) -2.00m/s^2.