An inline skater is traveling at a speed of 8.94 m/s; 120.0 s later on a level road, she is climbing an incline at a hill with a 15.09° angle and a speed of 7.15 m/s. (a) What is the change in her velocity? (b) What is her average acceleration during the 120 s time interval? Hint: V = 7.15 m/s, V0 = 8.94 m/s, θ = 15.09°, Δt = 120 s. Av = Vf – V0, tan θ = Avy/Avx.
The Correct Answer and Explanation is:
Let’s solve the problem step by step using vector analysis.
(a) Change in Velocity
We treat the skater’s velocity as a vector. Initially, she moves horizontally at 8.94 m/s. After 120 seconds, her velocity is 7.15 m/s at an upward angle of 15.09°. We resolve the final velocity into horizontal and vertical components:
- Final horizontal velocity: vfx=7.15cos(15.09∘)≈6.90 m/sv_{fx} = 7.15 \cos(15.09^\circ) ≈ 6.90 \, \text{m/s}
- Final vertical velocity: vfy=7.15sin(15.09∘)≈1.86 m/sv_{fy} = 7.15 \sin(15.09^\circ) ≈ 1.86 \, \text{m/s}
- Initial velocity vector: v⃗i=8.94 m/s\vec{v}_i = 8.94 \, \text{m/s} in horizontal direction vix=8.94 m/s,viy=0 m/sv_{ix} = 8.94 \, \text{m/s}, \quad v_{iy} = 0 \, \text{m/s}
- Change in velocity components: Δvx=vfx−vix=6.90−8.94=−2.04 m/s\Delta v_x = v_{fx} – v_{ix} = 6.90 – 8.94 = -2.04 \, \text{m/s} Δvy=vfy−viy=1.86−0=1.86 m/s\Delta v_y = v_{fy} – v_{iy} = 1.86 – 0 = 1.86 \, \text{m/s}
- Magnitude of change in velocity: ∣Δv⃗∣=(−2.04)2+(1.86)2≈2.75 m/s|\Delta \vec{v}| = \sqrt{(-2.04)^2 + (1.86)^2} ≈ 2.75 \, \text{m/s}
- Direction: ϕ=tan−1(1.862.04)≈42.3∘\phi = \tan^{-1}\left(\frac{1.86}{2.04}\right) ≈ 42.3^\circ above the horizontal, backward.
(b) Average Acceleration
Using the formula for average acceleration: a⃗avg=Δv⃗Δt=2.75 m/s120 s≈0.0229 m/s2\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{2.75 \, \text{m/s}}{120 \, \text{s}} ≈ 0.0229 \, \text{m/s}^2
Explanation
In this problem, the skater begins with a velocity purely in the horizontal direction. After some time, she starts climbing, which introduces a vertical component to her motion. Since velocity is a vector, both its magnitude and direction matter. That means a change in direction, even if speed drops only slightly, still results in a measurable change in velocity.
To compute how her motion evolved, we decomposed the final velocity into horizontal and vertical parts using trigonometric identities. The angle of inclination determined how much of the final speed contributed to vertical motion. The initial motion had no vertical component. So, introducing that vertical shift, even at a modest angle, produced a velocity change with both magnitude and direction.
From a physics standpoint, acceleration is the rate of change in velocity over time. The direction and magnitude of the velocity vector changed over 120 seconds, so we quantified this transformation with vector subtraction. That yielded a net velocity change of 2.75 m/s. Dividing this by the time interval gave us the average acceleration, which turned out to be a small number. This makes sense. Climbing a slope gradually slows the skater down and shifts her direction slightly upward. Since this transition is smooth and drawn out over time, the overall acceleration is gentle.
In essence, this exercise illustrates the importance of treating velocity as a vector and understanding how even subtle directional changes can influence overall motion. It also highlights how physics can model real-world scenarios like skating on an inclined path. Want to tackle another vector problem? I’m ready when you are.
