Challenge Problem Solutions: Two Dimensional Kinematics

[Pages:32]Two Dimensional Kinematics

Challenge Problem Solutions

Problem 1:

Suppose a MIT student wants to row across the Charles River. Suppose the water is moving downstream at a constant rate of 1.0 m/s. A second boat is floating downstream with the current. From the second boat's viewpoint, the student is rowing perpendicular to the current at 0.5 m/s. Suppose the river is 800 m wide.

a) What is the direction and magnitude of the velocity of the student as seen from an observer at rest along the bank of the river?

b) How far down river does the student land on the opposite bank?

c) How long does the student take to reach the other side?

Problem 1 Solutions: From observer on land

From observer on Boat A

Boat B moves at an angle with respect to the land.

Boat B moves horizontally to the current

!

vB

=

v!

!B

+

!

v

v!

= vA ^j , v!!B = vB! ?

!

vB

=

v!B ?+vA

^j

(( ) )( ) !

vB

=

v!B + vA2

1 2

vA = 1m " s#1 , v!B = 0.5m " s#1

(( ) ( ) ) !

vB

=

1

1m " s#1 2 + 0.5m " s#1 2 2 = 1.12m " s#1

tan$ # vA / v!B = 1.0m " s#1 / 0.5m " s#1 = 2

$ tan#1 (2) = 63.4"

Problem 2:

A person initially at rest throws a ball upward at an angle !0 with an initial speed v0 . He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval !t1 and then continues to run at a constant speed for a time interval !t2 . He catches the ball at exactly the same height he threw the ball. Let g be the gravitational constant. What was the person's acceleration a ?

Problem 2 Solutions:

In this problem there are two objects moving. The person and the ball. The ball undergoes projectile motion so we have the kinematic equations for the ball. The person undergoes two stages of motion. The first stage is constant acceleration and the second stage is constant velocity so we can write separate equations describing the position for each stage noting that the final position and velocity at the end of the first stage are the initial conditions at the beginning of the second stage. The constraint is that the ball and the person intersect at the end of the second stage. We first draw a coordinate system and a graph of the motion of the two objects. Let's choose the origin at the point the ball was released and assume that the person catches the ball at the same height above the ground as it was released.

The ball is in flight for a total time t f = !t1 + !t2 . The equations for the x and y positions of the ball are then

x2 (t) = v0 cos!0t ,

(2.1)

y2 (t )

=

v

0

sin!0t

"

1 2

gt 2

.

At the final instant t f = !t1 + !t2 , the ball is located at x2 (t f ) = v0 cos!0 ("t1 + "t2 )

Note that at t f = !t1 + !t2 ,

(2.2) (2.3)

0

=

y2 (t f

)

=

v 0 sin!0t f

"

1 2

gt f 2

which we can solve for the t f

t f

=

2v 0 sin!0 g

=

"t1

+ "t2

(2.4) (2.5)

This means that the initial speed and angle are related according to

!0

=

$ sin"1 &

%

g

(

#t1 + 2v 0

#t2

)

' ) (

v 0

=

g(!t1 + !t2 ) 2 sin "0

We will leave the rest of our results in terms of the angle !0 and v 0 .

(2.6) (2.7)

Stage 1: The equations for position and velocity of the person are:

x1 (t )

=

1 2

at 2 ,

vx1(t) = at

At the end of stage at t = !t1, the position and velocity of the person is given by

(2.8) (2.9)

x1 ( !t1 )

=

1 2

a(!t1)2 ,

vx1(!t1) = a!t1

(2.10) (2.11)

Stage 2: Let's reset our clock t = 0 when the first stage is done. Then initial position at

this

instant

is

x10

=

1 2

a(

!t1

)2

and

the

x-component of the

velocity at this

instant

is vx10 = a!t1 . The the equations for the position of the person for the second stage are then

x1 (t )

=

x10

+

vx01t

=

1 2

a( !t1 )2

+

a!t1t

(2.12)

In particular after an interval t = !t2 has elapsed the person is at the position

x1(!t2 )

=

1 2

a( !t1 )2

+

a!t1!t2

(2.13)

.

Since the person and the ball are at the same position, we can equate the right hand sides

of Eq. (2.3) and Eq. (2.13) and find that

v cos! ("t + "t ) = 1 a("t )2 + a"t "t

0

01

22

1

1 2

(2.14)

We can now solve Eq. (2.14) for the acceleration of the person

a

=

v 0

cos! ("t 01

+

"t ) 2

1 2

(

"t 1

)2

+

"t "t 1 2

(2.15)

Note that we could substitute Eq. (2.6) for the initial angle or Eq. (2.7) for the initial speed into Eq. (2.15).

Problem 3:

A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the shore. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at an acceleration that varies in time according to

ax = b1t

and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , b1 , h , and the gravitational constant g . You may neglect all air resistance.

Problem 3 Solution:

While the person is running, since the acceleration varies in time, in order to find the xcomponent of the velocity, we integrate the x-component of the acceleration with respect to time

t

t

" " vx (t) ! vx0 = 0 ax (t)dt = 0 (b1t )dt .

=

1 2

b1t 2

Since the runner started from rest vx0 = 0 and therefore

vx (t)

=

1 2

b1t 2

.

(3.1) (3.2)

We now integrate the x-component of the velocity with respect to time to find the displacement

t

t

" " x(t)

!

x0

=

0

vx (t)dt

=

0

(

1 2

b1t 2 )dt

.

=

1 6

b1t 3

(3.3)

Choose an origin from the point the runner began, so x0 = 0 , and the position function is

x(t )

=

1 6

b1t 3 .

(3.4)

Let t = t1 denote the instant the runner reaches the cliff, then

x(t1 )

=

d

=

1 6

b1t13

.

(3.5)

Hence

t1

=

! # %

6d b1

"1/ 3 $ &

.

(3.6)

The x-component of the velocity of the runner at t = t1 is tehn

vx (t1)

=

1 2

b1t12

=

1 2

b 1

! # %

6d b1

"2/3 $ . &

(3.7)

Let's restart our clock at t = 0 when the runner leaves the cliff. Let's pick our origin at the point of departure. Since the person is in free fall when in the air and we can neglect air resistance, the runner has acceleration ay = !g . So the y-component of the position of

the runner is given by

y(t )

=

y0

!

1 2

g

t 2

.

(3.8)

At t = 0 , y0 = 0 , so

y(t) = h ! 1 g t2 .

(3.9)

2

Let t = t2 denote the instant the runner hits the water, then

y(t2

)

=

!h

=

!

1 2

g

t 2 2

.

(3.10)

So we can solve for the time of flight,

t2

=

! # %

2h g

"1/ 2 $ &

.

(3.11)

The horizontal position of the runner from the edge of the cliff is given by x(t) = vx0t .

(3.12)

Since we know the runner leaves the cliff with an x-component of the velocity equal to the x-component of the velocity of the runner at t = t1 , we have that

x(t )

=

1 2

b 1

! #%

6d b1

"2/3 $ &

t

.

(3.13)

At t = t2 , the horizontal distance to the rock is then

x(t2 )

=

s

=

1 2

b1

! # %

6d b 1

"2/3 $ &

t 2

=

1 2

b1

! % #

6d b1

"2/3 $ &

! #%

2h g

"1/ 2 $ &

.

(3.14)

or

s

=

1 2

b1

! # %

6d b1

"2/3 $&

! 2h #%

g

"1/ 2 $&

.

(3.15)

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