Physics Problem Set: Calculating Distance, Speed, and Acceleration, Slides of Physics

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Practice Midterm #1 Solutions

Physics 6A

1. You drive your car at a speed of 40 km/hr for 1 hour, then slow down to 30 km/hr for the next 20 km. How fardid you drive, and what was your average speed?

20 km

40 km

We can draw a simple diagram withdistances and times. For each section weuse the formula “Distance = Speed x Time”

hr

t

t

km

Section

km

hr 1

x

Section

2

2

hr km

hr km

1

1 hr

2/3 hr

The

total

distance

is

60km.

Total

time

is

hr.

Divide

these

to

get

the

average

speed:

hr km

53

kmhr

speed .

avg

answer

(b)

3. A rock is thrown upward from the top of a 300-meter high cliff. The initial speed is 10 m/s. If we ignore airresistance, how long does it take for the rock to hit the ground below?

y=

v y=300m

0

=10m/s

First we can draw a sketch and choose a coordinate system. I havechosen the ground as y=0. That makes the initial height y

0

= 300m.

We also know the initial velocity, and the final height. We can use oneof our basic kinematic equations:

(

) (

sec

t

sec

t

t

t

t

t

t

m

t g t v y y

or

2

2

2

m s

1 2

s m

2

12

0

0

2

This gives us a quadratic equation, whichwe can solve with the quadratic formula.

a

2

ac

4

b

b

t

2

Choose the positive value. Answer (b).

4. Two rocks are thrown with the same initial speed from the top of a 300-meter high cliff. Rock A is thrownupward and Rock B is thrown downward. If we ignore air resistance, which of the following is true about thespeeds of the rocks just before they hit the ground? a)

Rock A is faster.

b)

Rock B is faster.

c)

Rock A and Rock B have equal speeds.

d)

There is not enough information to determine the answer.

y=

y=300m

Rock

A

y=

y=300m

Rock

B

The rocks will both land at the same

speed

. Rock A goes up to some high point, then turns around. By the time it

passes the starting point on the way down, it is moving at the initial speed, just opposite direction. So it is just likerock B at this point. Of course rock B hits first, but they are both moving the same speed when they hit. Answer (c)

6. An airplane is flying East at a speed of 300 km/hr. A gust of wind blows in a Northwest direction at 50 km/hr.Find the new speed and direction of the plane.

x

y

300

50

v

total

Northwest means halfwaybetween North and West

We need to add the vectors together, so first we must find their components:

(

)

(

)

hr km

hr km

y ,

wind

hr km

hr km

x ,

wind

hr km

y ,

plane

hr km

x ,

plane

sin

v

cos

v

v

v

Now add the components together, x with x and y with y.

hr km

hr km

hr km

y ,

total

hr km

hr km

hr km

x ,

total

v

v

Finally, we can put these components together using the Pythagorean Theorem:

hr km

2

2

total

v

r

We can calculate the angle from our right-triangle rules, and look at our picture tosee that the plane is flying a little bit North of East.

° = θ ⇒ = θ

6 .

7

tan

Answer (c)

7. Bob has 2 pet rocks. Their names are Elvis and Pedro. Bob drops Pedro from a high bridge to the river below.When Pedro has fallen 4m, Bob drops Elvis. As the rocks continue their free-fall, what happens to their separationdistance?a) The distance increases. b) The distance decreases. c) The distance stays the same. d) There is not enough information to determine the answer. Think about the speeds of the rocks. Since Pedro gets a head start, he is already moving when Elvis starts to fall.So Pedro is moving faster than Elvis, and they both continue to accelerate at the same rate (gravity). Both speedskeep increasing, but Pedro is always faster. Thus Pedro moves farther than Elvis during each second of their fall.So as they fall, the rocks get farther apart. Answer (a)

9. A soccer ball is kicked at an angle of 30

above the horizontal on a level field. The ball lands 45m from where it

was kicked. What was the initial speed of the ball? Again, start with a picture. The initial velocity willbreak down into components as follows:

0

45

v

0

v

ox

v

oy

θ

θ

θ

sin

v

v

cos

v

v

0

y , 0

0

x , 0

This is a projectile problem, so we use ourprojectile formulas:

a

v

v

t

v

x

x

x

x , 0

x

x ,

0

0

0

2

y , 0

2

y , 0

2

12

y ,

0

0

y y g 2 v v

gt

v

v

gt

t

v

y

y

Similar to the previous problem, we can find the time from the y equation, then plug into the x-equation.

g

sin

v

2

t

t

sin

v

gt

gt

t

sin

v

gt

t

v

y

y

0

0

2

1 2

2

1 2

0

2

1 2

y ,

0

0

θ

θ

− ⋅ θ ⋅ + =

s m

0

m s

(^20)

(^20)

s m

0

0 x , 0

0

v

cos

sin

m

v

cos

sin

v

2

m

g

sin

v

2

cos

v

m

t

v

x

x

2

2

θ ⋅ ⋅ θ ⋅ + =

Answer (c)

10. A golf ball is launched at an angle with an initial speed of 30 m/s, at an angle of 50

above the horizontal. The

ball lands on a green that is 5m above where the ball was struck. How far has the ball traveled in the horizontaldirection by the time it lands?

5m

V

0

=

X =?

First draw a picture, then break the initialvelocity into components:

(

)

(

)

s m

s m

y , 0

s m

s m

x , 0

sin

v

cos

v

Again we will use projectile equations:

a

v

v

t

v

x

x

x

x , 0

x

x , 0

0

0

2

y , 0

2

y , 0

2

12

y ,

0

0

y y g 2 v v

gt

v

v

gt

t

v

y

y

Using y

0

=0 and y=5m in the first vertical equation, we can find the time:

(

)

s

t

s

t

t

t

9 .

4

t

t

m

5

gt

t

v

y

y

or

2

2

s m

12

s m

2

1 2

y , 0

0

2

a

2

ac

4

b

b

t

2

Quadratic formula:

The ball is at height 5m twice – once on the way up, and then again on the way down. We want the 4.46s time.

(

) (

m

s

x

t

v

x

x

s m

x ,

0

0

Answer (a)

12. A jet plane comes in for a downward dive as shown in Figure 3.39. The bottom part of the path is a quartercircle having a radius of curvature of 350m. According to medical tests, pilots lose consciousness at anacceleration of 5.5g. At what speed will the pilot black out for this dive? The given acceleration is 5.5g. This means 5.5 times theacceleration due to gravity:

2

2

m s

m s

350m

The path of the airplane is circular, so the given acceleration mustbe centripetal (toward the center). We have a formula for centripetal acceleration:

s m

2

s m

2

cent

v

m

v

r v

a

2

Answer (a)

13. Two children are on a merry-go-round which spins at a rate of 1 revolution every 5 seconds. Bobby isstanding at a position that is 0.75m from the center, and Cindy is 1.50m from the center. Which statement is trueabout their linear speeds? a) Their speeds are equal b)

Bobby’s speed is twice Cindy’s speed.

c)

Cindy’s speed is one-and-a-half times Bobby’s speed.

d) Cindy’s speed is twice Bobby’s speed.

0.75m

1.5m

We don’t need to do a lot of calculations for this one.Enough information is given to find the speed of eachchild, but that is too much effort. Understanding therelationship between angular speed and linear speed willmake this problem seem easy. Both children are on the same rotating object, so theymust have the same

angular

speed (they rotate at the

same rate). However, Cindy is farther from the center, soshe has to travel a larger distance every time she goesaround. That is, her

linear

speed must be greater. Since

she is twice as far from the center, she has to travel twiceas far for each rotation, thus her speed is twice as fast. Answer (d) Here is the plug-into-the-equations solution (you need to know the circumference of a circle is 2

π

r):

s m

Cindy

s m

Bobby

s

5

m

5 .

1

v

s

5

m

v

time

ce

tan

dis

v

π

π

v v

s m s m

Bobby Cindy