Question

a boat entered a marina with an initial velocity of 2.58 m/s [W 25.0 N]. over an interval of 4.00 s the captain turned the boat towards a dock while they slowed the boat to a final velocity of 1.15 m/s. what was the average acceleration of the boat during the parking sequence?

Answer 1

Answer on Question #73561, Physics – Mechanics, Relativity

Question:

A boat entered a marina with an initial velocity of $2.58\ \mathrm{m/s}$ [W 25.0 N]. over an interval of 4.00 s the captain turned the boat towards a dock while they slowed the boat to a final velocity of 1.15 m/s. what was the average acceleration of the boat during the parking sequence?

Solution:

Using formula below:

$$v = v_0 + a \cdot t$$

where $v = 1.15\ \mathrm{m/s}$, $v_0 = 2.58\ \mathrm{m/s}$, $t = 4\ \mathrm{s}$, we got:

$$a = \frac{v - v_0}{t} = \frac{1.15 - 2.58}{4} = -0.3575\ (\mathrm{m/s^2})$$

Answer: $a = -0.3575\ \mathrm{m/s^2}$

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