Physics CH2

A car is driven from point B to point C and then driven to point A. What is thedisplacement of the car?

-5 km

A polar bear starts at the North Pole. It travels 1.0 km South, then 0.5 km East, then 1.0 kmNorth, then 0.5km West to return to its starting point. This trip takes 30 minutes. What was the bear's average velocity?

0 km

A polar bear starts at the North Pole. It travels 1.0 km South, then 0.5 km East, then 1.0 kmNorth, then 0.5km West to return to its starting point. This trip takes 30 minutes.What was the bear's displacement?

0 km

A car is driven from point B to point A. What is thedistance travelled?

5 km

a light year is the distance that light travels in one year. The speed of light is 3.00 x 10^8 m/s. How many miles are there in one light year? (1 mi= 1609m, 1 y =365 d)

5.88 x 10^12 mi

From a height of 300 m, a ball is thrown horizontally with an initial speed of 25 m/s.How far does it travel horizontally in the first 2 seconds?

50 m

If you are driving 25 m/s along a straight road and you look to the side for 3.0 s, how fardo you travel during this inattentive period?

75 m

A racquetball strikes a wall with a speed of 30 m/s and rebounds with a speed of 26 m/s. The collision takes 20ms. What is the average acceleration of the ball during collision?

Although the change in speed is 4 m/s, the change in velocityis actually 30m/s + 26m/s = 56m/s because the ball also changesdirection. The change in time is given as 20ms (0.02s). aave= ?v / ?t = (56m/s) / (0.02s) =2800m/s

Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the way you plan to stop for dinner. If the fastest you can safely drive is 65 mi/h, what is the longest time you can spend over dinner and still arrive just in time for the meeting?

Answer: 1.55 h Explanation: Let the time to spend over dinner be t. Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive. Speed = distance/time The distance is 549 and maximum speed is 65 mi/h.

The captain orders his starship to accelerate from rest at a rate of "1 g" (1 g = 9.8 m/s2). How many days does it take the starship to reach 5% the speed of light? (Light travels at 3.0 × 108 m/s.)

Answer: 17.7 days Explanation: Since the ship accelerates from the rest, its initial velocity would be equal to 0. So, Acceleration of the star-ship = a = 1 g = 9.8 m/s² We need to find how many days will it take the ship to reach 5% of the speed of light. Speed of light is m/s. 5% of the speed of light = m/s This means, the final velocity of the star-ship will be: We have the initial velocity, final velocity and the acceleration. We need to find the time(t). First equation of motion relates these quantities as: Using the values in this equation, we get: Thus, the star-ship will take 1,530,612.245 seconds to reach to 5% the speed of light. Now we need to convert this time to days. Since, there are 60 seconds in a minute: 1,530,612.245 seconds = minutes Since, there are 60 minutes in an hour: 25,510.20 minutes = hours Since, there are 24 hours in a day: 425.17 hours = days Thus, it will take approximately 17.7 days (approximately 17 days and 17 hours) to reach to 5% the speed of light

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant acceleration. The car comes to a stop in a distance of 120 m.1st Find Acceleration -> then use it to solve.How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

Answer: a = -2.82 m/s² v = 18.4 m/s Explanation: Initial velocity, u = 26.0 m/s Final velocity, v = 0 m/s Distance travelled, s = 120m Using v² = u² + 2as 0 = 26² + 2(120)a a= -2.82 m/s² Velocity of the car when it was 60.0m past the point where the brakes were applied: u = 26.0 m/s a = -2.82 m/s² s = 60m Using v² = u² + 2as v² = 26² + 2(-2.82)(60) v² = 337.6 v = 18.4 m/s

The captain orders his starship to accelerate from rest at a rate of "1 g" (1 g = 9.8 m/s2). How many days does it take the starship to reach 10% the speed of light? (Light travels at 3.0 × 108 m/s.)

Answer:35days Explanation: acceleration a=velocity v/time a is. given to be 9.8m/s2 V=0.1x speed of light,=3.0x107 t =3.0x10*7/9.8 = 3.06x10*6sec converting to days I. E divide by 86400 will make t to be 35.days to the nearest significant number.

A car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? a car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? 112 m 292 m 4.57 m 146 m 9.14 m

The car will travel up to a distance of 146.28 m, hence the correct answer is 146 m. Since the car is decelerating with the constant acceleration, so we can apply the third equation of motion. v^2=u^2-2aS here, v is the final speed of the car, which is 0 as the car stops, u is the initial speed=32 m/s a is the constant acceleration=3.5 m/s^2 and S is the distance cover by the car before it stops. Now plugging the values in the third equation of motion 0=32^2-(2*3.5*S) S=146.28 m Therefore the car will cover a distance of 146 m before it stops.

Suppose that an object travels from one point in space to another. Make a comparison between the magnitude of displacement and the distance by this object.

The displacement is either less than or equal to the distance traveled.

The graph above is the position-versus time graph of an object. Which of thefollowing is true regarding the motion of the object?

The object is moving in the negative direction & The speed of the object is decreasing

Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of its displacement, but the magnitude of the displacement can never be greater than the distance it runs.

True

A car slows down from 30 m/s to rest in a distance of 85 m. What is its acceleration, assumed constant?

a = (v² - v₀²) ÷ [2(x - x₀)] a = [(0)² - (30 m/s)²] ÷ [2(85 m - 0 m)] a = (-900 m²/s²) ÷ (170 m) a = -5.3 m/s²

The kinematic equation for finding the average velocity at constant acceleration?

avg. v = (v + v₀) ÷ 2

The kinematic equation for finding the velocity of an object after any elapsed time?

v = v₀ + at

A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?

vf2 - vi2 = 2ax vf = final speed after traveling a distance x at constant acceleration; vf = 42 m/s vi = initial speed = 0 (starts from rest) a = constant acceleration while traveling a distance x x = distance traveled at constant acceleration; x = 5.6 km = 5600 m a = vf2/(2x) = 422/(2*5600) m/s2 = 0.1575 m/s2 ≅ 0.16 m/s2

The kinematic equation useful in situations where time (t) is not known?

v² = v₀² + 2a(x - x₀)

A light plane must reach a speed of 32 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 2.8 m/s²?

v² = v₀² + 2a(x - x₀) (x - x₀) = (v² - v₀²) ÷ (2a) x = (32 m/s - 0 m/s) ÷(2 × 2.8 m/s²) x = 182.85 m x = 190 m

The kinematic equation for finding the position (x) of an object after a time (t) at constant acceleration?

x = x₀ + (avg. v)t

The kinematic equation for finding time given distance and acceleration (constant)?

x = x₀ + v₀t + ½at²