Over a time interval of 1.52 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.5 km/s to -15.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. (a) Number 27752 I Inits

Over a time interval of 1.52 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.5 km/s to -15.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. (a) Number 27752 I Inits

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter3: Motion Along A Straight Line

Section: Chapter Questions

Problem 9CQ: If you divide the total distance traveled on a car trip (as determined by the odometer) by the...

See similar textbooks

Similar questions

16. Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.9 km/s to – 18.5 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s²) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
An adventure racer ran 1.7 km north, then 3.2 km 63° to the south of due east, then 1.4 km west. She completed the trip in 2 hr: 5 min :8 s. What was the magnitude of her average velocity over the entire trip in m/s? Express your answer in metres/second to two decimal places (do not include units in your answer).
Over a time interval of 1.52 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.5 km/s to -15.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s²) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. (a) Number i (b) Number i Units Units
Over a time interval of 1.52 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.5 km/s to -15.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s²) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. (a) Number i -36.3 (b) Number i eTextbook and Media Save for Later Units Units No units S kg m m/s m/s^2 N days Attempts: 0 of 3 used Submit Answer
Over a time interval of 1.54 years, the velocity of a planet orbiting a distant star reverses direction, changing from +16.3 km/s to -19.3 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. (a) Number i (b) Numberi Units Units
"A jet plane is cruising at 320 m/sm/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.8 km , the jet is moving with a speed of 400 m/sm/s. What is the jet's acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units." I am having issues setting up word problems to begin solving. Need an easier method or technique to help deciper these.
In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5190 kmkm away, and released. It found its way back to its nest 13.5 daysdays after release. If we place the origin in the nest and extend the +x+x-axis to the release point, what was the bird's average velocity for the return flight? Express your answer in meters per second. vav.−xvav.−x = ________________ m/s What was the bird's average velocity for the whole episode, from leaving the nest to returning? Express your answer in meters per second. vav.vav. = __________________ m/s
An object is moving with an initial velocity of 5.8 i m/s and a final velocity 13.8 i m/s. The time taken for acceleration is 2.2 seconds. Calculate the acceleration in m/s2 to 2 sf. In your answer, you do not enter the i unit vector, but you do need to enter negative signs, if appropriate. e.g. if your answer is a = -5.4 i m/s,2 you would enter -5.4
A kicker kicks a football upward from the ground at an initial velocity of 63 feet per second. The height of the football stadium is 70 feet. The height an object reaches with respect to time is modeled by the following equation: gr +vt +s In the equation, g is -32 ft/sec', v is the initial velocity, s is the initial height, and t is time in seconds. Write a function that models this situation as related to the number of seconds since kickoff. 1. 2. Sketch and describe the graph of this function, including intercepts and maximum height. 3. At what times is the football the same height as the stadium? Explain your answer. 4. Suppose the initial velocity of the kicked football is 68 feet per second. At what times is the football the same height as the top of the stadium? Justify your answer. Now consider that the kicker is trying to kick an extra point. A linebacker on the opposing team has a maximum reach of 10 feet, which includes his height, full extension of his arms, and his…
the rocket starts from rest at t=0 and travel straight up. it's height above the ground as a function of time can be approximated by s=bt^2+ct^3, where b and c are constants. at t=10s, the rocket's velocity and acceleration are v= 229m/s and a=28.2 m/s^2 A.) determine the time in seconds at which the rocket reaches supersonic speed (325 m/s) B.) determine the value of constant B C.) determine the value of constant C
My question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2. That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2. My question is why you get a different result doing it the first way than you get doing it the second way.
You are driving home from school steadily at 97 km/hkm/h for 160 kmkm . It then begins to rain and you slow to 61 km/hkm/h instantly. You arrive home after driving 4.5 hours. How far is your hometown from school? answer using two significant figures. What was your average speed? answer using two significant figures.