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)

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

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

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach