Question

A ladybug is moving at a speed of 1 cm/s. She begins to decelerate at a rate of –0.25 cm/s2until she comes to rest. How much distance does the ladybug cover from the moment she begins to decelerate until the moment she comes to rest? (HINT: This is a two-part problem. First, use an equation to determine the time. Then, plug this value into another equation to determine the distance.)

Answer 1

The distance covered by the ladybug from the moment she begins to de-accelerate until the moment she comes to rest is 2 cm.

What are the three equations of motion?

The three equations of motion are -

v = u + at

S = ut + 1/2 at²

v² - u² = 2aS

Given is a ladybug which is moving at a speed of 1 cm/s. She begins to decelerate at a rate of -0.25 cm/s² until she comes to rest. From this we can write -

initial velocity [u] = 1 cm/s

final velocity [v] = 0 m/s

acceleration [a] = -0.25 cm/s² = 1/4 cm/s²

Using the first equation of motion -

v = u + at

0 = 1 - t/4

t/4 = 1

t = 4 sec

Using second equation of motion -

S = ut + 1/2 at²

S = 1 x 4 - 0.5 x 0.25 x 4 x 4

S = 4 - 2

S = 2 cm

Alternative method

Using third equation of motion, we can directly find out the distance covered -

v² - u² = 2aS

- 1 = - 2 x 1/4 x S

S/2 = 1

S = 2 cm

Therefore, the distance covered by the ladybug from the moment she begins to de-accelerate until the moment she comes to rest is 2 cm.

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