Question

A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed to 1.89 m/s. He covers 17.5 m until he lands on shore. What was his acceleration?

Answer 1

Answer:

0.085m/sec²

Explanation:

We will use the equation

${v}^{2} - u^{2} = 2as \\ v = 1.89 \\ u = 0.78 \\ s = 17.5 \\ a = ? \\ {1.89}^{2} - {.78}^{2} = 2 \times a \times 17.5 \\ 2.9637 = 2 \times a \times 17.5 \\ a = \frac{2.9637}{2 \times 17.5} = 0.085$

Answer 2

Answer:

0.085

Explanation:

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