Question

An object moves with constant acceleration 4.00m/s² and over a time interval reaches a final velocity of 12m/s.

a) If the original velocity is 6.00m/s, what is its displacement during the time interval?
b) What is the distance it travels during this interval?
c) If its original velocity is −6.00m/s, what is its displacement during this interval?
d) What is the total distance it travels during the interval in part (c)?

Answer 1

a) The displacement during the time interval is 18 m. b) The distance traveled during this interval is also 18 m. c) When the initial velocity is -6 m/s, the displacement during the time interval is 54 m. d) The total distance traveled during the interval in part (c) is 72 m.

Step-by-step explanation:

a) In order to find the displacement, we can use the equation of motion: Final velocity squared (v^2) = initial velocity squared (u^2) + 2 times acceleration (a) times displacement (s).

Rearranging the equation, we have: displacement (s) = (v^2 - u^2)/(2a).

Plugging in the given values, we get: displacement (s) = (12^2 - 6^2)/(2 * 4) = 18 m.

b) The distance traveled during this interval can be found by taking the absolute value of the displacement.

So, the distance traveled is 18 m.

c) Using the same equation of motion, we can find the displacement when the initial velocity is -6 m/s.

Plugging in the given values, we get: displacement (s) = (12^2 - (-6)^2)/(2 * 4) = 54 m.

d) The total distance traveled during the interval in part (c) is equal to the sum of the absolute values of the initial and final displacements.

So, the total distance traveled is 54 m + 18 m = 72 m.