Question

A package slides across a frictionless horizontal surface with constant acceleration 4.05 m/s² and over a time interval reaches a final velocity of 11.2 m/s. (a) If its initial velocity is 5.60 m/s, what is its displacement (in m) during the time interval? (Indicate the direction with the sign of your answer.) __________ .

Answer 1

Sure, I'd be happy to explain this!

First, let's figure out how much time passed during this time interval.

We will use the first equation of motion to determine the time. The equation is:

v = u + at

where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time

By rearranging this formula, we can get the expression to calculate the time:

t = (v - u)/a

Substituting in the given values:

t = (11.2 m/s - 5.60 m/s) / 4.05 m/s²

For these given values, we get t ≈ 1.3827 seconds

Now, let's calculate the displacement of the package during the time interval.

To do this, we can use a different equation of motion:

s = ut + 0.5at²

where:
- s is the displacement
- u is the initial velocity
- t is the time
- a is the acceleration

Substituting in the given values and the time we just calculated:

s = (5.60 m/s * 1.3827s) + (0.5 * 4.05 m/s² * (1.3827s)²)

For these given values and calculated time, we find that s ≈ 11.615 m

Therefore, the package's displacement over this time interval is approximately 11.615 meters.