Question

A bowling ball starts from rest and moves 300 m down a long, downwardly angled track in 22.4 sec.a. What is its speed at the end of the track? [27 m/s]b. What is its acceleration? [l.2 m/s2]c. Sketch three graphs: Δx vs. Δt, v vs. Δt and a vs. Δt. We are interested in the initial and final values, and the general shape of the graph, (linear, exponential, etc)

Answer 1

The speed at the end of the track = 27 m/s

The acceleration = 1.2 m/s²

Please find the Δx vs Δt, v vs Δt, a vs Δt

Step-by-step explanation:

We have;

x = u·t + 1/2·a·t²

Where;

x = The distance = 300 m

u = The initial velocity = 0 m/s (Ball at rest)

t = The time taken = 22.4 s

Therefore;

300 = 0 + 1/2×a×22.4²

a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²

v = u + a×t

∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s

Part of the table of values is as follows;

t, x, v

0, 0, 0

0.4, 0.095663, 0.478316

0.8, 0.382653, 0.956632

1.2, 0.860969, 1.434948

1.6, 1.530611, 1.913264

2, 2.39158, 2.39158

2.4, 3.443875, 2.869896

2.8, 4.687497, 3.348212

3.2, 6.122445, 3.826528

3.6, 7.748719, 4.304844