Physics Chapter 2 Motion

V=0 a=9.8 m/s^2 down

A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path

v=0,a=9.8m/s2down.

A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?

To solve for initial velocity v1= Vo + gt 0 = Vo - (9.8) (1.6) Vo= 15.7 m/s To solve for height d= vi * t + 0.5*g*t2 d= 0(1.6) + (0.5)(9.80)(1.6)^2 d= 12.5m

A ball player catches a ball 3.20 s after throwing it vertically upward. With what speed did he throw it? What height did it reach?

If its acceleration (or deceleration) was uniform, then its average speed during that time was 25.0/2, or 12.5 meters per second. At that average speed, in 8.60 seconds it traveled 12.5 times 8.60, or 107.5 meters.

A car slows down uniformly from a speed of 25.0 m/s to rest in 8.60 s . How far did it travel in that time?

The car's average acceleration points due west

A car traveling due east at 20 m/s reverses its direction over a period of 10 seconds so that it is now traveling due west at 20 m/s. What is the direction of the car's average acceleration over this period?

16d

On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters during the first t seconds of its fall. Assuming the rock does not hit the ground first, how far will it fall during the first 4t seconds of its fall?

The truck travels twice as far as the car.

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car?

4V

On the Moon, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock has a speed v after falling from its point of release for a time t. Assuming that the rock does not hit the ground first, how fast will it be moving after it has fallen for a time 4t from its point of release?

The rock has a downward acceleration of 9.8 m/s2.

Suppose that you toss a rock upward so that it rises and then falls back to the earth. If the acceleration due to gravity is 9.8 m/sec2, what is the rock's acceleration at the instant that it reaches the top of its trajectory (where its velocity is momentarily zero)? Assume that air resistance is negligible.

The acceleration of the car is positive and decreasing.

The figure shows the velocity versus time graph for a car driving on a straight road. Describe the acceleration of the car.

The position versus time graph is parabolic.

What shape do we expect for a graph showing the position of a motorcycle versus time if the motorcycle is speeding up with a constant acceleration?

Velocity decreasing below the x-axis. Why?

Which figure could represent the velocity versus time graph of a motorcycle whose speed is increasing?

Average acceleration is always equal to the change in velocity of an object divided by the time interval

Which of the following best describes how to calculate the average acceleration of any object?

If release occurs at t=0, the vertical height ot the package after that, in meters, is y = 145 + 5.19 t -4.9 t^2 (The 4.9 is g/2, and g is the acceleration of gravity in units of m/s^2) Solve the equation y = 0 and you will get the time that it his the ground. It is a quadratic equation with two roots. Take the positive one. t = [-5.19 -sqrt(5.19^2 +4*145*4.9)]/(-9.8)= seconds

A helicopter is ascending vertically with a speed of 5.19 m/s . At a height of 145 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v0 for the package?]

vf2=vo2+2ax vf2−vo2=2ax 2ax=vf2−vo2 x=vf2−vo2/2a x=(35ms)^2−(0ms)^2/2(3.2ms2) x= 190m Round down

A light plane must reach a speed of 35 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.2 m/s2?

average speed = total distance/total time Convert the total time from minutes to seconds. Keeping in mind that 1 min = 60 s, we have t= 15.4 min * 60 s/min = 924s Each lap is 400 m, and the person completes 8 laps, so the total distance covered by the person is S= 8 * 400m = 3200m The average speed is just the ratio between the total distance and the total time: v= s/t = 3200m/924s = 3.46 m/s

A person jogs eight complete laps around a 400-m track in a total time of 15.4 min. Calculate the average speed, in m/s.

The average velocity is equal to the ratio between the total displacement of the person and the total time: v= d/t however, the total displacement is zero, because the person returns to his starting point (the track is a closed circuit). Therefore, d=0 and the average velocity is zero as well: v=0

A person jogs eight complete laps around a 400-m track in a total time of 15.4 min. Calculate the average velocity, in m/s.

Average velocity = (x2 - x1)/(t2-t1) = -3.4 cm/s

A rolling ball moves from x1 = 8.3 cm to x2 = -4.5 cm during the time from t1 = 3.1 s to t2 = 6.9 s. What is its average velocity over this time interval?

What is her acceleration in m/s2? a= v/t a= 9.50/1.24 a= 7.66 m/s^2 What is her acceleration in km/h2? 12960*7.66= 99273.6= 9.93x10^4

A sprinter accelerates from rest to 9.50 m/s in 1.24 s . What is her acceleration in m/s2? What is her acceleration in km/h2?

Vo = 0 m/s Do = 0 m a = 9.8m/s^2 (gravity) t = 3.25s To find distance, we use (D= Do + Vo x t + 1/2 x a x t^2) D = 0 + 0 + (1/2)(9.8)(3.48^2) D = 59.3 m

A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.48 s . How high is the cliff?

The object will slow down, momentarily stopping, then pick up speed moving to the left.

At time t=0 an object is traveling to the right along the +x axis at a speed of 10.0 m/s with acceleration -2.0 m/s2 . Which statement is true?

A car that is traveling in the -x direction, decreasing in speed

In which of the following cases does a car have a negative velocity and a positive acceleration?