2 physics
Which of the following provides an example of one-dimensional motion in which average speed is not equal to the magnitude of average velocity? A boy rides a bicycle along a straight path from his house to the market. A pitcher throws a baseball along a perfectly straight line. A man throws a ball vertically upwards and catches it when it falls back. A girl walks 80m north, halts for a while, and walks 100m further north to reach home.
A man throws a ball vertically upwards and catches it when it falls back. The net displacement of the ball is zero. So the average velocity will be zero. But there has been distance traveled by the ball, so the average speed will not be zero.
You are throwing a ball straight up in the air. Describe the ball's velocity and acceleration at the highest point.
At the highest point, acceleration is nonzero, but its velocity is zero.
Identify some advantages of metric units.
Conversion between units is easier in metric units bc base 10
Acceleration is calculated using velocity. True or False: Acceleration is a scalar quantity.
False
True or False: For a ball thrown straight up in the air, the magnitude of the average velocity, for any time interval, is always greater than the magnitude of its initial velocity.
False
Is it possible to have a description of motion without a frame of reference?
No, because motion is relative. It is not possible to describe the motion of the body without a frame of reference. Motion is relative and is always described with respect to a chosen frame of reference.
True or False: Total distance traveled must be either equal to or greater than magnitude of total displacement.
True
Two runners travel along the same straight path. They start at the same time, and they end at the same time, but at the halfway mark, they have different instantaneous velocities. Is it possible for them to have the same average velocity for the trip?
Yes, because average velocity depends on the net or total displacement. Instantaneous velocities might change, but they do not have a bearing on the average velocity as it is the net displacement divided by total time.
During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
200m-(1.50s)v=(16.30s)v v=200m/(16.30s + 1.50s)=11.2 m/s
Which of the following provides a correct answer for a problem that can be solved using the kinematic equations?
A body starts from rest and accelerates at 4m/s^2 for 2s. The body's final velocity is 8m/s. It is a kinematic problem. The final velocity of the body can be obtained by using the kinematic equations.
A cyclist rides 3km west and then turns around and rides 2km east. What is her displacement? What distance does she ride? What is the magnitude of her displacement?
Displacement is −1km, distance is 5km, and the magnitude of displacement is 1km1km. Displacement is defined as the shortest distance between the initial and final positions along with the direction, and distance is defined as the total path covered by the object.
Why is acceleration a vector quantity?
It is a vector quantity because it has magnitude as well as direction.
Which of the following statements explains why a racecar going around a curve is accelerating, even if the speed is constant?
The car is accelerating because the direction of velocity is changing.
If a logarithmic scale is used on a graph, what can you assume about the range of the data?
The data extend over a wide range of values. The logarithmic scale comes to the rescue when the data extends over a wide range of values.
Comparing Distance Traveled with Displacement: A Subway Train What are the distances traveled for the motions of the train
The displacement was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km. The displacement was −1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.
A student calculated the final velocity of a train that decelerated from 30.5m/s and got an answer of −43.34m/s. Which of the following might indicate that he made a mistake in his calculation?
The sign of the final velocity is wrong. The magnitude of the answer is larger than the magnitude of the original velocity.
True or false? The net displacement in a velocity vs. time graph is the area under the curve. The value if net displacement in the position vs time graph is not different from the net displacement calculated through the velocity vs. time graph.
True
What is the difference between distance and displacement? Describe the differences between vectors and scalars using physical quantities as examples.
distance is a scalar and it has no direction attached to it, whereas displacement is a vector and direction is important. Scalar quantities have only magnitude but no direction, whereas vector quantities have magnitude as well as direction.
A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?
initial velocity is 0, final velocity of 30 m/s, t is 7.00 seconds to change from 0 to 30 meters per second. a=change in velocity / change in time (30 m/s - 0 m/s) / 7 s = 4.29 m/s^2
A police officer uses a device to clock a vehicle traveling 85miles per hr (mph). What property is the officer's device measuring?
instantaneous speed
A world record was set for the men's 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.
the speed is constant for 6.69s so x2=vt=(6.69s)v=100m-x1 accelerate for 3.00s x1=1/2at^2=1/2vt=(1.50s)v 100m-(1.50s)v=(6.69s)v v=100m/(6.69s + 1.50s) v=12.2 m/s a=v/t a=12.2m/s/3s=4.07m/s^2
If a particle is accelerating, which value cannot be constant?
velocity Remember that acceleration is defined as the rate of change of velocity; velocity is a vector, so the change in velocity must either be a change in magnitude (speed) or a change in direction.
Calculating Deceleration: The Subway Train Given 𝑣0=−20 km/h, 𝑣f=0 km/h, Δ𝑡=10.0 s. What is its average acceleration?
Δ𝑣=𝑣f−𝑣0=0−(−20 km/h)=+20 km/h 𝑎=Δ𝑣/Δ𝑡=+20.0 km/h/10.0 s 𝑎=(+20.0 km/h/10.0 s)(10^3m/1 km)(1 h/3600 s)=+0.556 m/s^2
Calculate Acceleration: A Subway Train Slowing Down Given 𝑣0=30.0 km/h, 𝑣f=0 km/h (the train is stopped, so its velocity is 0), and Δ𝑡=8.00 s. What is its average acceleration while stopping?
Δ𝑣=𝑣f−𝑣0=0−30.0 km/h=−30.0 km/h 𝑎=Δ𝑣/Δ𝑡=−30.0 km/h/8.00 s 𝑎=Δ𝑣/Δ𝑡=(−30.0 km/h/8.00 s)(10^3m/1 km)(1 h/3600 s)=−1.04 m/s^2
Calculating Displacement: A Subway Train What are the magnitude and sign of displacements for the motions of the subway train given 𝑥f=6.70 km and 𝑥0=4.70 km and 𝑥′f=3.75 km and 𝑥′0=5.25 km
Δ𝑥=𝑥f−𝑥0=6.70 km−4.70 km=+2.00 km displacement Δ𝑥′=𝑥′f−𝑥′0=3.75 km−5.25 km=−1.50 km displacement
A driver pulls into a parking space at a velocity of 1.6𝑚/𝑠 and brakes to a stop in 0.8s. What is the car's rate of acceleration?
−2.00m/s2 The car is decelerating and is defined as rate of change of velocity with time.
Calculating Acceleration: A Subway Train Speeding Up Given 𝑣0=0(the trains starts at rest), 𝑣f=30.0 km/h and Δ𝑡=20.0s What is its average acceleration during that time interval?
𝑎=Δ𝑣/Δ𝑡=+30.0 km/h/20.0 s 𝑎=(+30 km/h20.0 s)(10^3m/1 km)(1 h/3600 s)=0.417 m/s^2
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?
𝑎=Δ𝑣/Δ𝑡=−15.0 m/s/1.80 s=−8.33 m/s^2.
Calculating Average Velocity: The Subway Train Given 𝑥′f=3.75 km, 𝑥′0=5.25 km, Δ𝑡=5.00 min. What is the average velocity of the train if it takes 5.00 min to make its trip?
𝑣=Δ𝑥′/Δ𝑡=−1.50 km/5.00 min 𝑣=Δ𝑥′/Δ𝑡=(−1.50 km/5.00 min)(60 min/1 h)=−18.0 km/h
An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s^2 given 𝑣0=70.0 m/s 𝑎=−1.50 m/s^2, 𝑡=40.0s.
𝑣=𝑣0+𝑎𝑡=70.0 m/s+(−1.50 m/s^2)(40.0 s)=10.0 m/s
Calculating Final Velocity: Dragsters Dragsters can achieve average accelerations of 26.0 m/s^2. Suppose such a dragster accelerates from rest at this rate for 5.56 s. Given 𝑣0=0, a is 26.0m/s^2, x-x0=402 m
𝑣^2=0+2(26.0 m/s^2)(402 m) 𝑣^2=2.09×10^4m^2/s^2 𝑣=sqrt (2.09×10^4m^2/s^2)=145 m/s.
On dry concrete, a car can decelerate at a rate of 7.00 m/s^2, whereas on wet concrete it can decelerate at only 5.00 m/s^2. Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
𝑣^2=𝑣0^2+2𝑎(𝑥−𝑥0) 𝑥−𝑥0=(𝑣^2−𝑣0^2)/2𝑎 𝑥−0=(0^2−(30.0 m/s)^2)/2(−7.00 m/s^2) 𝑥=64.3 m on dry concrete The only difference is that the deceleration is -5.00 m/s^2 The result is 𝑥wet=90.0 m on wet concrete 𝑣=30.0 m/s; 𝑡reaction=0.500s; 𝑎reaction=0. We take 𝑥0−reaction to be 0. We are looking for 𝑥reaction. 𝑥=𝑥0+𝑣𝑡 𝑥=0+(30.0 m/s)(0.500 s)=15.0 m 64.3 m + 15.0 m = 79.3 m when dry 90.0 m + 15.0 m = 105 m when wet
Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s^2, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)
𝑥=𝑥0+𝑣0𝑡+1/2𝑎𝑡^2 200 m=0 m+(10.0 m/s)𝑡+1/2(2.00 m/s^2)𝑡^2 𝑡^2+10𝑡−200=0 𝑡=10.0sand−20.0s t=10 s
Dragsters can achieve average accelerations of 26.0 m/s^2. Suppose such a dragster accelerates from rest at this rate for 5.56 s. Given 𝑣0=0, a is 26.0m/s^2 and 𝑡 is 5.56 s. How far does it travel in this time?
𝑥=𝑥0+𝑣0𝑡+1/2𝑎𝑡^2 𝑥=1/2𝑎𝑡^2 𝑥=1/2(26.0 m/s^2)(5.56 s)^2 x=402 m
A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero? Given 𝑣=4.00 m/s, Δ𝑡=2.00 min, x0=0 m
𝑥=𝑥0+𝑣𝑡=0+(4.00 m/s)(120 s)=480 m