Question

p25. a particle moves along the x axis according to the equation $x = 2.00 + 3.00t - 1.00t^2$ 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.

Explanation:

Step1: Substitute t=3.00s into position equation

$x = 2.00 + 3.00(3.00) - 1.00(3.00)^2$

Step2: Calculate position value

$x = 2.00 + 9.00 - 9.00 = 2.00$

Step3: Derive velocity from position

$v = \frac{dx}{dt} = 3.00 - 2.00t$

Step4: Substitute t=3.00s into velocity equation

$v = 3.00 - 2.00(3.00)$

Step5: Calculate velocity value

$v = 3.00 - 6.00 = -3.00$

Step6: Derive acceleration from velocity

$a = \frac{dv}{dt} = -2.00$

Step7: Acceleration is constant, no substitution needed

$a = -2.00$

Answer:

(a) 2.00 meters
(b) -3.00 m/s
(c) -2.00 m/s²