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Physics - PowerPoint Presentation

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Physics - PPT Presentation

Force and Motion Test Question Explanations Objective To review the questions contained in your test to understand how to solve any questions that you may have missed The question number listed here are arbitrary and do not reflect any particular test questions were randomized ID: 449100

train question 112 line question train line 112 part time minutes car speed find 300 stop distance travel velocity

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Slide1

Physics

Force and Motion Test

Question ExplanationsSlide2

Objective

To review the questions contained in your test to understand how to solve any questions that you may have missed

The question number listed here are arbitrary and do not reflect any particular test (questions were randomized)Slide3

Question 1

Usain

Bolt ran the 200 m in 19.19 s. What is the average speed of his run

?

All we need for this problems is v = d/t

200 m/ 19.19s = 10.42 m/sSlide4

Question 2

A bird's egg falls from a nest that is 18 m above the ground. How long will it take that egg to hit

the ground

(Neglect air resistance

)?

We know that y= 18m

We also know that a = 9.81m/s

2

AND that vi = 0 the instant it starts to fall

We are trying to find time

Let’s see: y, a, vi, and t

y =

v

i

t

+ ½ at

2

18 = 0 + ½ (9.81)t

2

18 = 4.905t

2

18/4.905 = t

2

t = 1.92 sSlide5

Question 3

A boy is trying to throw a rose to a girl he likes who lives on the second floor (about 8m high).

How fast

will he need to throw the flower straight up so that its highest point is the same as the height

of the

second floor

?

We know that y = 8

We know that

v

f

= 0 since it stops momentarily at the top of its height

We know that a = -9.81 m/s

2

(since gravity is in the opposite direction of travel

We need to find v

i

Let’s see: y,

v

f

, a, v

i

v

f

2

= v

i

2

+ 2aySlide6

Question 3, continued

0 = v

i

2

+ 2(-9.81)(8)

0 = v

i

2

+ -156.96

156.96 = v

i

2

v

i

= 12.5 m/sSlide7

Question 4

A car starts from rest and accelerates at a constant rate in a straight line. In the first second, the

car covers

a distance of 2.0 meters. How much additional distance will the car cover during

the second

second of its motion

?

For the first second, we know

x = 2 m, vi = 0 m/s, t = 1 s

It says that the car is accelerating but not by how much. Let’s find out

x

=

v

i

t

+ ½ at

2

2 = ½ a (1)

2

2 = ½ a

a = 4 m/s

2Slide8

Question 4, continued

Since it is accelerating, it is going faster and faster

It DOES NOT cover the same distance in each second

How far does it travel from start to the 2s mark?

x = ?, a = 4 m/s2, vi = 0 m/s, t = 2s

x =

v

i

t

+ ½ at

2

x = 0 + ½ (4) (2)

2

x = 8 m

Since it would have covered 2 m in the first second, then it would have covered 6 m in the second

second

.Slide9

Question 5

Carl Lewis set a world record for the 100.0 m run with a time of 9.86 s. If, after reaching the

finish line

, Mr. Lewis walked directly back to his starting point in 90.9 s, what is the magnitude of

his average

velocity for the 200.0 m

?

This checks for understanding of concepts

If he went 100m and then came back 100m, then his displacement is zero

Since velocity is displacement divided by time, then average velocity was zeroSlide10

Question 6

What

is the velocity of the object at t = 7.0 s?

The velocity is equal to the slope of the line at that time

m = rise/run

 (

0-40)/(

6-5)  -

40/1

= -

40 m/sSlide11

Question 7

At a drag race, a car accelerates for 2.6 s to a speed of 112 m/s. Maintaining that top speed,

it reaches

the finish line 400 m from the starting line. Once there, a parachute deploys enacting

an acceleration

of 39 m/s

2

.

A) What is the acceleration of the car when the race starts?

B) How much time does it take the car to reach the finish line?

C) How much time does it take to stop?

D) How far from the starting line does the car finally stop?Slide12

Question 7 part A

We know that: t= 2.6 s,

v

f

= 112 m/s

We also know that vi = 0 m/s

We can use the formula a = (

v

f

– v

i

) / t

a = (112 – 0)/ 2.6

a = 43 m/s2 Slide13

Question 7 part B

It accelerates in the first 2.6 s

It doesn’t stop there, it keeps travelling forward BUT NOT accelerating after 2.6s

How far did it travel in the first 2.6s?

x =

v

i

t

+ ½ at

2

x = 0 + ½ (43)(2.6)2

x = 145 m

The finish line is 400 m from the starting line

400 – 145 = 255 m left to go

It is travelling at a constant speed (112 m/s)

v = d/t

 t = d/v  255 / 112 = 2.3 s

The entire trip would have taken 2.6 + 2.3 = 4.9sSlide14

Question 7 part C

Once it crosses the line, the parachute slows it down

For that portion of the trip, we know:

a = -39 m/s

2

(because it is slowing down), v

i

= 112 m/s, and

v

f

= 0 m/s

We are trying to find time

Once again, we can use: a = (

v

f

– v

i

) / t

-39 = (0 – 112)/t

-39 t = -112t = 2.9sSlide15

Question 7 part D

We already know that it travelled 400m to get to the finish line

Now we need to know how much farther it travelled

Once again, we know that: a = -39 m/s

2

, v

i

= 112 m/s, and

v

f

= 0 m/s

We need to find x

v

f

2

= v

i

2

+ 2ax

0 = 112

2

+ 2(-39)x

0 = 12544 -78x

-12544 = -78 x

x = 161 m

Total distance 400 + 161 = 561 mSlide16

Question 8

A runaway train is barreling down the tracks at a speed of 75 km/h compared to the

tracks. A

solution has been devised to send another train down a parallel track to catch up and stop

the train

. The second locomotive will not pull cargo and can achieve a speed of 90 km/h compared

to the

track. The first train has a 5 km head start. Once the train catches up, it will take 15 minutes

to slow

the train to a stop

A) How fast would the second train travel compared to the first?

B) How long will it take the second train to catch up to the second in minutes?

C) What would the deceleration of first train be to slow to a stop in 15 minutes.

D) What is the total distance that the first train will travel after being caught up to by the second?Slide17

Question 8 part A

Compared to each other, one train is travelling 15 km/hSlide18

Question 8 part B

We know that there is a 5 km head start

Since the relative speed of the 2

nd

train is 15 km/h compared to the first, you can work the problem as if the first train were not moving

v = d/t

 t = d/v  5/15 = 1/3 of an hour

The question asks for minutes: 1 hour/3 times 60 minutes/hour = 20 minutesSlide19

Question 8 part C

When the “target” train starts to slow down, we know:

v

i

= 75 km/h, t = 15 minutes,

v

f

= 0 km/h

Since, v

i

is in km/h and time is in minutes, you will have to convert one or the other (or both) so that they have the same units

NOTE: for your answer key, I made 3 different versions of this same answer

One method would be to change time to hours: t=.25 hours

We can use: a = (

v

f

– v

i

) / t

a = (0 – 75)/.25 = - 300 km/h

2Slide20

Question 8 part D

What we know

v

i

= 75 km/h; a = -300 km/h

2

,

v

f

= 0 km/h

We are trying to find x

v

f

2

= v

i

2

+ 2ax

0 = 75

2

+ 2(-300)x

0 = 5625 -600x

-5625 = -300 x

x = 9.4 kmSlide21

Question part D, alternate

You may have decided to use time instead

v

i

= 75 km/h, a = - 300 km/h

2

, t = .25 h

x =

v

i

t

+ ½ at

2

x = (75)(.25) + ½ (-300)(.25)

2

x = 18.75 – 9.375

x = 9.4 km