Solutions: Motion Questions Answered

[terpsgirl]

Motion questions...

Need a little help...

If the sun is a distance of 1.5 x 10^8 km from earth, how long does it take sunlight to reach Earth if it moves at 3.0 X 10^8 m/s?

A rocket moves through out space at 11,000 m/s. At this rate how much time would be required to travel the distance from the Earth to Moon, which is 38,000 Km?

A rifle is fired straight up, the bullet leaves the rifle with an initial velocity magnitude of 724 m/s. After 5.000 s the velocity is 675 m/s. At what rate is the bullet decelerated?

THX

 

[AKG]

speed = distance/time. Use this equation to answer the first two questions (you may have to rearrange it to isolate for different variables). For constant acceleration (or "deceleration") as it is in your case:

acceleration = (velocity at time2 - velocity at time1)/(time2 -time1)

and (time2 - time1) is the change in time, or the time elapsed.

 

[M98Ranger]

!

For the first question, we can use the formula d = vt, where d is the distance, v is the velocity, and t is the time. We can rearrange the formula to solve for t: t = d/v. Plugging in the given values, we get t = (1.5 x 10^8 km) / (3.0 x 10^8 m/s) = 0.5 seconds. So it takes 0.5 seconds for sunlight to reach Earth from the sun.

For the second question, we can use the same formula, d = vt, but this time we are solving for t. Plugging in the given values, we get t = (38,000 km) / (11,000 m/s) = 3.45 seconds. So it would take 3.45 seconds for the rocket to travel from Earth to the Moon.

For the third question, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. We can rearrange the formula to solve for a: a = (v-u)/t. Plugging in the given values, we get a = (675 m/s - 724 m/s) / 5.000 s = -9.8 m/s^2. This means that the bullet is decelerating at a rate of 9.8 m/s^2.