Choices B and C are wrong because an object moving in a circular path at constant speed is constantly accelerating.)

10mi

4h 2 mi/h

¹₂

17. C By definition of acceleration,

a v

= ∆

t

= − = ⋅

⁻

⋅

⁻

∆

(60 0)mi/h

6s 10 mi h s

^{1 1}

18. D Using Big Five #3, with v

₀

= 0 and converting hours to seconds, we find

∆x =

at

=

⋅







 ⋅









1 2

2

1

2 10mi

h s

1h

60 60s (6s)(66s) 1 20 mi

=

19. C The object’s velocity is a constant v = –3 m/s, so using Big Five #1 (with

v v

= ), we find

∆x = v∆t = (–3 m/s)(4 s) = –12 m Thus, x – x

₀

= –12 m, so x = –8 m since x

₀

= 4 m.

20. D Average speed is defined as total (as opposed to net) distance divided by time. Since 500 meters was covered in 25 seconds, the horse’s average speed was (500 m)/(25 s) = 20 m/s.

21. A By completing one lap around the track, the horse’s final position coincided with his initial posi-tion. Thus, the displacement was zero, which implies the average velocity was also zero.

22. D Velocity is speed plus direction. Only choice D contains such data. (Note: Choice A is a dis-tance, choice B is a displacement, and choice C is a speed.)

23. D By definition, acceleration implies a change in velocity. (Choice A is wrong because an object can accelerate without changing its direction [just drive in a straight line and step on the gas].

Choices B and C are wrong because an object moving in a circular path at constant speed is constantly accelerating.)

24. B Assuming that the time delay between the lightning flash and its visual reception is essentially zero, the distance to the lightning is calculated by determining the distance traveled by the sound of the accompanying thunder. This distance is d = vt = (340 m/s)(5 s) = 1700 m.

25. D If the velocity during the period of time in question were constant, then statements A and C

would be false (since the average velocity would equal the final velocity). And, if the initial and

final velocities were different, then statement B would be false.

26. B Since the round-trip time is 4 seconds, the time it takes the rock to reach the top of its path (where v = 0) is half this: 2 seconds. Then, applying the equation a = ∆v / ∆t to the first half of the trip, we find a = (0 – 3.2)/2 = –1.6 m/s

²

, so the magnitude of the moon’s gravitational accel-eration is 1.6 m/s

²

.

27. C The horizontal component of the initial velocity is v

_0x

= v

₀

cos θ, and the vertical component is

v_0y

= v

₀

sin θ. Therefore, as θ increases from 0 to 90°, v

_0x

decreases (because cos θ decreases) and

v_0y

increases (because sin θ increases).

28. C The horizontal component of the initial velocity is v

_0x

= v

₀

cos θ which is proportional to cos θ.

29. A Since the projectile experiences no horizontal acceleration, there will be no change in the hori-zontal component of its velocity. This eliminates choices C and D. On its way down, however, the projectile’s vertical speed increases, since it experiences the downward acceleration due to earth’s gravity.

30. D The horizontal distance traveled by a projectile—a quantity called its range—is equal to its hori-zontal velocity, v

_0x

, multiplied by its total flight time (assuming that the take-off and landing points are at the same horizontal level). The total flight time is equal to twice the time necessary to reach the top of its path, which is where its vertical velocity drops to zero. The equation ∆v

_y

=

g∆t gives ∆t = v_0y

/g as the time to reach the topmost point, so the total flight time is twice this:

2v

_0y

/g. Thus, the projectile’s range is R = 2v

_0xv_0y

/g. Since R depends on both v

_0x

and v

_0y

, there is insufficient information given here to decide which projectile will have the greater range.

31. D Ignoring the miniscule variation in the gravitational acceleration with altitude—assuming that the altitude is small compared to the earth’s radius—the acceleration of the projectile is con-stant during its entire flight.

32. B Using Big Five #2, we find

∆v = a∆t = (2 m/s

²

)(5 s) = 10 m/s

Since v – v

₀

= 10 m/s and v

₀

= 6 m/s, we must have v = 16 m/s.

33. C Notice that the actual value of a is neither given nor asked for; we therefore use Big Five #1:

∆

x v t

= ∆ =

¹₂

(

v₀

+

v t

) ∆ =

¹₂

( 4 14 3 + )( ) = 27 m

34. D Since v

₀

= 0, Big Five #5 becomes v

²

= 2a∆x. Since a is a constant, this equation tells us that

∆x is proportional to v

²

. So, if the final velocity v is increased by a factor of 4, then ∆x must in-crease by a factor of 4

²

= 16.

35. C Using Big Five #5 and solving for a, we get

v v a x v v a

v a a

2 0

2 1

2 0 2

0

2 1

8 3

4 0

2 1

4

2 2

= + ⇒ ( ) ^{= +}

⇒ − = ⇒

∆ ( )

== −3

v₀²

36. B Apply Big Five #3:

∆

x v t

=

₀

∆ +

¹₂a t

( ) ∆

²

= ( )( ) 10 4 +

¹₂

( )( ) 2 4

²

= 56 m

37. B By definition of average velocity,

v

= ∆ ∆

x

/

t

= (56m)/(4s) 14m/s =

38. D Car #1 will catch up to Car #2 when the two cars have the same position. Therefore, we must find the positions of the cars and set them equal to each other. The position of Car #2 is easy to calculate since the speed is constant: Using distance equals rate times time, we find x

₂

= 20t. For Car #1 we use Big Five #3, with v

₀

= 0 and a = 5/8: we get x

₁

= (5/16)t

²

. Setting x

₁

and x

₂

equal to each other and solving for t, we find that t = 64 seconds.

Another solution involves finding how long it takes for Car #1 to reach the speed of Car #2. Us-ing v = at, we get 20 = (5/8)t, so t = 32 seconds. Since it takes that long just for Car #1 to reach Car #2’s speed, it will take even more time to actually catch up to Car #2’s position, so the only possibility among the given choices is choice D.

39. D Let’s try Big Five #3, with ∆x = x (since x

₀

= 0):

x v t

=

₀

∆ +

¹₂a t

( ) ∆

²

= ( )( )

v₀

3 + −

¹₂

( )( ) 2 3

²

= 3

v₀

− 9

Since v

₀

is negative (because we’re told that the object is traveling in the –x direction at the beginning of the time interval), x must be less than –9. The only possibility among the given choices is choice D.

40. A The object started at x = 0 at time t = 0, moved to position x = 2 two seconds later, then just sat there. Evidently, it only traveled 2 m. (The distance traveled by an object is given by the area under the velocity vs. time graph, not the area under the position vs. time graph.)

41. A The acceleration is the slope of the v vs. t graph. The slope of the segment from O to P is 2/1 = 2, and the slope of the segment from P to Q = –2/4 = –1/2. Therefore, the ratio of the magnitudes of these accelerations, a

_PQ

: a

_OP

, is 1/2 : 2 = 1:4.

42. B The area under the v vs. t graph will give the distance traveled by the object. In this case, we have a triangle with base 5 and height 2, and the area of this triangle is

1

2

(base)(height) =

¹₂

(5 s)(2 m/s) = 5 m

43. C We can see that no velocity graphed is negative, so it’s OK here to say speed = velocity. Since a is constant between t = 1 and t = 5 (since the v vs. t graph is a straight line here), the average speed is just the average of the initial and final speeds. Therefore,

v

=

¹

(

v₀

+ =

v

)

¹

(

v_att=1

+

v_att=5

) =

¹

( 2 0 + = ) 1 m/s

44. D

Statement I is false: the object returned to its original velocity at t = 5, not to its original posi-tion. In fact, because it always had a positive velocity, the object was always moving to the

right. It never had a negative velocity, so it could never have backed up and returned to its original position. Notice that recognizing that Statement I is false eliminates two of the an-swers, choices A and B. Statement III is false, also. As mentioned in the discussion of Statement I, the object’s velocity was always positive. Since the velocity never changed sign, the object’s direction of travel never changed either. The correct answer must be choice D. Statement II is also false: the average speed between t = 0 and t = 1 is 1 m/s, as is the average speed between

t = 1 and t = 5. See the calculation we did in the solution to #43 above.

45. B Call “down” positive, so a = g is positive. Then using Big Five #3 with v

₀

= 0, we find

∆ ∆ ∆

x a t t x

= 1 ⇒ =

a

= ≈

2

2 3

( )

2

2(128ft) sec

32ft/s

²

46. C Use Big Five #5, we see that choice C is the best answer since

v² v₀² a x v a x

5 7 12

2 2 2 32 128

2 2 2 2 2 2

= + ⇒ = = ⋅ ⋅ =

⋅ ⋅ = ⋅ =

∆ ∆

66

2 64 1 5 ≈ ( . ) = 96 ft/s

47. A Once released, the motion of the arrow is due to gravity alone, and the acceleration due to grav-ity—the vector g—always points down.

48. B One solution uses Big Five #5. All we have to realize is that at the top of its (purely vertical) flight, its velocity v = 0. If we call “up” the positive direction, then the acceleration a = –g is negative, and we find

v v a x x v v

a

v v

2 0

2 2

02 2

02 02

2 2

0

2 2

= + ⇒ = − = −

− =

∆ ∆

(

g

)

g

49. B We use Big Five #3 throughout this solution (calling “down” positive, so a = g is positive). Dur-ing the first second, the object falls a distance of

¹₂a t

( ) ∆

²

=

¹₂g

(1)

²

=

¹₂g

. During the first two seconds, it falls a distance of

¹₂a t

( ) ∆

²

=

¹₂g

(2)

²

= 2

g

. Therefore, the distance it falls just during the 2

^nd

second is 2

g

−

¹₂g

=

³₂g

, so the desired ratio is (

³₂g

)/ (

¹₂g

) ^{= 3 .}

50. B Let’s decide to call “down” positive, so a = g is positive, and use Big Five #3 (with v

₀

= 0):

∆

x

=

¹₂a t

( ) ∆

²

=

¹₂g

( ) ∆

t ²

=

¹₂

( )( ) 10 3

²

= 45 m

51. A Call “down” positive; then a = g is positive, and the initial velocity v

₀

is –8 m/s (because the initial velocity is “up,” and “up” is the negative direction). Big Five #3 then tells us

∆

x v t

=

0

∆ +

1a t

∆ = − ( ) ^{+ = − + =}

2

2 1

2

8 3 10 3

2

24 45

( ) ( ) ( )( ) 221m

52. A Call “down” positive; then a = g is positive. Now use Big Five #5:

v v a x x v v

a

2 0 2

2

02 2 2

2

2

14 0 2 9 8

196 19 6

= +

= −

= − =

∆

∆ ( . ) . == 10m

53. B First of all, eliminate choice A. T must be greater than t, so the ratio T/t must be greater than 1.

We’re going to call “down” positive, so a = g is positive, and use Big Five #3 with v

₀

= 0. Let ∆x be the entire distance of 980 m. From ∆x =

¹₂aT²

=

¹₂gT²

we getT = 2∆ / g

x

. The time t to fall the first

¹

2

∆x is found from

¹₂

∆x =

¹₂at²

= g

¹₂ t²

, sot = ∆ / g

x

. Thus, the desired ratio is

In document TPR Science WB 2011 (Page 143-147)