Question

required information
a particle has a constant acceleration of 5.00 m/s² to the east. at time t = 0, it is 2.00 m east of the origin and its velocity is 20.0 m/s north. take +x as east and +y as north.
what is the y - component of its position vector at t = 2.40 s?
□ m north

Explanation:

Step1: Identify y-direction motion parameters

Acceleration in y-direction $a_y = 0$ (no north/south acceleration), initial y-position $y_0 = 0$, initial y-velocity $v_{0y} = 20.0\ \text{m/s}$, time $t=2.40\ \text{s}$.

Step2: Apply position formula for y-axis

Use constant velocity position equation: $y = y_0 + v_{0y}t$
Substitute values: $y = 0 + (20.0\ \text{m/s})(2.40\ \text{s})$

Step3: Calculate final y-position

$y = 20.0 \times 2.40 = 48.0$

Answer:

$48.0$ m north