Question

2) A boat travels 12.0 m while it reduces its velocity from 9.5 m/s to 5.5 m/s. What is the
boat's acceleration while it travels that distance?

Answer 1

Answer:

2.5m/s^2

Explanation:

Note that the boat is reducing its speed. It is having negative acceleration or deceleration.

V^2 = u^2 -2as ( minus sign is used because of speed is reduced)

Given that,

s = 12 m

v = 5.5 m/s

u = 9.5 m/s

a = (v^2 - u^2) ÷ (-2s)

a= ( 5.5^2 - 9.5^2) ÷ ( -2× 12)

a = 2.5 m/s^2

Answer 2

The acceleration of the boat is -2.5 m/s²

The given paramters;

distance traveled by the boat, d = 12 m

initial velocity of the boat, u = 9.5 m/s

final velocity of the boat, v = 5.5 m/s

The acceleration of the boat is calculated as;

$v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(5.5)^2 -(9.5)^2 }{2(12)} \\\\a = -2.5 \ m/s^2$

Thus, the deceleration of the boat is 2.5 m/s²

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