Quiz Solutions for Physics 2211 - Spring 2008 - Problem 1, Quizzes of Physics

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Physics 2211 Quiz #1 Solutions Spring 2008

Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected.

I. (16 points) A car is travelling in a straight line with velocity +v 0. At time t = 0, it begins to accelerate at a = +k

√

t, where k is a positive constant. Through what distance does the car travel between times t = 0 and t = T? Express this distance in terms of any or all of v 0 , k, T , and physical or mathematical constants.

....................... The acceleration is the time-derivative of the velocity, so

a = dv/dt ⇒ v =

∫

a dt =

∫

kt^1 /^2 dt = k

t^3 /^2 3 / 2

+ C

where C is an integration constant. At time t = 0, the velocity is v 0 , so

v 0 = 23 k

(

03 /^2

)

• C ⇒ C = v 0 so v = 23 kt^3 /^2 + v 0

Since the velocity is the time-derivative of the position, the displacement will be the definite integral of the velocity over time.

v = dx/dt ⇒ ∆x =

∫ (^) t 2

t 1

v dt =

∫ T

0

(

2 3 kt

3 / (^2) + v 0

)

dt = 23 k t^5 /^2 5 / 2

• v 0 t

∣

∣

∣

∣

T

0

So

∆x = 154 k T 5 /^2 + v 0 T

II. (16 points) You are motorcycling home, when you find your way blocked by a swollen stream. At that instant, you are directly north of your home, and a distance D away, as illustrated. You travel a distance a along the stream, in a direction θ south of east, to a bridge. How far from your home is the bridge, in terms of any or all of D, a, θ, and physical or mathematical constants? (The bridge has negligible length, so it does not matter which side you are on.)

....................... Let the vector D~ point from the starting position to home. Let the vector ~a point from the starting position to the bridge. Let the vector ~b point from the bridge to home. Then ~a + ~b = D~. Looking at the East components, and noting that the East component of D~ is zero

aE + bE = DE = 0 ⇒ bE = −aE = −a cos θ

Looking next at the North components, note that the North component of D~ is −D.

aN + bN = DN = −D

So

bN = −D − aN = −D − (−a sin θ) = a sin θ − D

The magnitude of ~b can be found from its components with the Pythagorean Theorem ∣ ∣ ∣~b

∣

∣

∣ =

√

b^2 E + b^2 N =

√

(−a cos θ)^2 + (a sin θ − D)^2 or

√

(a cos θ)^2 + (a sin θ − D)^2

1. (6 points) In the problem above, let the distance you’ve determined be b. The displacement vector from the bridge to your home, then, is ~b. The vector ~b has North and East components bN and bE , respectively. In what direction should you travel to get to your home from the bridge? Let the positive East axis be zero, with positive angles toward the North. ....................... With the positive East axis defined as zero and positive angles toward the North, one can think of the East axis as the x axis, and the North axis as the y axis. The angle, then, would be tan−^1 (bN /bE ), except that the East component of ~b is negative (from the diagram), so the arctangent is off by 180◦^ or π radians.

tan−^1

(

bN bE

)

• π

III. (16 points) Two cars, A and B, are waiting side by side at a red light. When the light turns green at time t = 0 s, Car A heads down the road with a speed that increases 1.0 m/s each second for 4.0 s, and 2.0 m/s each second after that. The driver of Car B isn’t paying attention, and doesn’t start until 4.0 s after the light turns green. However, the speed of Car B increases 3.0 m/s each second. Sketch a quantitatively correct speed vs. time graph for each car. Be sure to label the curves. From your graph, at the instant the cars have the same speed, which car is ahead, and by how much distance?

.......................

The curves cross at 8.0 s, which is the instant the cars have the same speed of 12 m/s. The distance each car has travelled is the area under its curve from its starting time to 8.0 s. The area under curve B is a triangle with base 4.0 s and height 12 m/s.

xB = 12 (4.0 s × 12 m/s) = 24 m

The area under curve A can be calculated in three pieces. There is a triangle with base 4.0 s and height 4 m/s, a rectangle with base 4.0 s and height 4 m/s, and a triangle with base 4.0 s and height 8 m/s,

xA = 12 (4.0 s × 4 m/s) + (4.0 s × 4 m/s) + 12 (4.0 s × 8 m/s) = 8 m + 16 m + 16 m = 40 m

Therefore, car A is ahead by 40 m - 24 m = 16 m.

2. (6 points) In the problem above, what must be true about the curves you’ve drawn at the instant the cars have the same speed? ....................... Since this is a graph of speed as a function of time, when the cars have the same speed... The two curves must have the same value at that instant.

3. (10 points) Consider two vectors, A~ and B~. Let their sum be S~ = A~ + B~ and their difference be D~ = A~ − B~. How must the magnitudes of S~ and D~ be related? ....................... Imagine two simple cases. First, if A~ and B~ point in the same direction, the magnitude of their sum will be greater than the magnitude of their difference. On the other hand, if A~ and B~ point in opposite directions, the magnitude of their sum will be less than the magnitude of their difference. Therefore, there can be no necessary relationship between the sum and difference of two general vectors such as A~ and B~ in this question, and of the choices provided... None of the others is correct.
4. (10 points) Under what conditions, if any, might an object traveling with constant speed have a non-zero acceleration? ....................... The acceleration cannot be in the same direction as the velocity, or the object’s speed would increase. Nor can the acceleration be opposite the velocity, as the object’s speed would decrease. Further, the acceleration can have no component parallel or anti-parallel to the velocity if the object is to have constant speed. The acceleration can only be perpendicular to the velocity, which will change the direction of the velocity. The direction of the acceleration must change as well, then, if it is to remain perpendicular to the new velocity. Since the acceleration is changing direction, it is not constant. The acceleration must be perpendicular to the velocity and cannot be constant.

5. (10 points) The ball rolls up the ramp, then back down. The positive direction has been defined as down the ramp. Which is the appropriate graph of acceleration vs. time? (On Earth.)

.......................

The ball has a constant acceleration down the slope. With that choice of coordinate system, down the slope is positive.

6. (10 points) The graph shows the acceleration of an object moving in a straight line as a function of time. The velocity of the object is positive at time t = 0. During the time represented on the graph, the speed of this object... ....................... This acceleration is always positive. Since the velocity is positive as well, the acceleration is in the same direction as the velocity and the speed of this object increases.