The distance covered by the ladybug from the moment she begins to de-accelerate until the moment she comes to rest is 2 cm.
<h3>What are the three equations of motion?</h3>
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
<h3>
Alternative method</h3>
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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