Question

2)An athlete starts at point A and runs at a constant speed of 6.60 m/s around a round track 250 m in diameter.Find the x and y components of this runner’s average velocity and average acceleration between points A and B.

Answer 1

Answer 2

Answer:

Given:

• Diameter = 250 m → Radius $r=125\text{m}$.

• Speed $v=6.60\text{m/s}$.

• Points A and B are on opposite ends of the track (half-circle apart).

1. Average Velocity (${\vec{v}}_{\text{avg}}$):

• Displacement (A→B): Straight-line distance = $2r=250\text{m}$ (x-direction).

• Time ($t$): Half-circumference distance = $\pi r$, so $t=\frac{\pi r}{v}$.

• Components:

  $$v_{\text{avg, x}}=\frac{\mathrm{\Delta }x}{t}=\frac{250}{\pi \times 125\mathrm{/}6.60}=\frac{250\times 6.60}{125\pi }\approx 4.20\text{m/s}(x)$$

  $v_{\text{avg, y}}=0\text{m/s}$ (no net vertical displacement).

2. Average Acceleration (${\vec{a}}_{\text{avg}}$):

• Velocity change ($\mathrm{\Delta }\vec{v}$):
  At A: ${\vec{v}}_A=+6.60\text{m/s}(y)$.
  At B: ${\vec{v}}_B=-6.60\text{m/s}(y)$.
  $\mathrm{\Delta }\vec{v}={\vec{v}}_B-{\vec{v}}_A=-13.2\text{m/s}(y)$.

• 
  

• Components:

  $$a_{\text{avg, y}}=\frac{\mathrm{\Delta }{v}_y}{t}=\frac{-13.2}{\pi \times 125\mathrm{/}6.60}\approx -0.221{\text{m/s}}^2(y)$$

  $a_{\text{avg, x}}=0{\text{m/s}}^2$.