Solution to Problems: Kinematics P69: 18, 25, 49, 58, 66, 90 G
18. The position of a particle as a function of time is given by r = ( 7.60t ,8.85, − t 2 ) m . Determine the particle's velocity and acceleration as a function of time. Solution: The velocity is the first derivative of the position with respect to time: G
G
v = dr dt = ( 7.60, 0, −2t ) m/s . And the acceleration is the first derivative of the velocity with respect to time: G
G
a = dv dt = ( 0, 0, −2 ) m/s 2 . G
25. The position of a particle is given by r = ( 6.0cos3.0t ,6.0sin 3.0t ) meters. Determine (a) the velocity vector, v, and (b) the acceleration vector, a. (c) What is the path of this particle? (d) What is the relation between r and a (given a formula), and G
G
between r and a (give an angle)? (e) Show that a = v 2 / r . G
G
Solution: (a) the velocity can be obtained as v = dr dt = ( −18sin 3.0t ,18cos3.0t ) m/s , G
G
(b) the acceleration can be obtained as a = dv dt = ( −54cos3.0t , −54sin 3.0t ) m/s2 . (c) From the problem, it can be verified that x 2 + y 2 = 6.02 m 2 , so the path is a circle. (d) From the equations, it is easily to find that a / r = 9 s −2 ; and the direction of acceleration is opposite to that of position. (e)From the expressions above, v 2 / r = (182 / 6 ) m/s 2 = 54 m/s 2 = a, QED. 49. A person stands at the base of a hill that is a straight incline making an 55° angle to the horizontal. For a given initial speed v0 , at what angle should objects be thrown so that the distance d they land up the hill is as large as possible? Solution: It is a problem of a projectile motion. Assuming the angle is θ, then the
horizontal displacement is x = v0t cosθ , and the vertical displacement
1
2
is y = v0t sinθ − gt
2
(the initial position is chosen as the origin). While the object
lands up the hill, one has y = x tan ϕ . From above 3 equations, one can get for given
2v0 cosθ
( tan θ − tan ϕ ) . And the
g
v0 , θ and ϕ , the object lands up the hill at t =
distance they land up the hill thus
is d = x 2 + y 2 = x (1 + tan 2 ϕ ) =
From
∂d 2v0
=
∂θ
g
2v02
g
(1 + tan ϕ ) cos θ ( tan θ − tan ϕ ) ⎜⎝⎛ ϕ ≤ θ < 12 π ⎟⎠⎞ ,
2
2
(1 + tan ϕ ) ( cos 2θ + sin 2θ tan ϕ ) = 0, one gets θ = ϕ2 + π4
2
. And the maximum distance is d max =
=
2v02
g
2
0
(1 + tan ϕ ) cos θ ( tan θ − tan ϕ ) = vg ( sin 2θ − 2 cos θ tan ϕ ) (1 + tan ϕ )
2
2
v02
( cos ϕ − tan ϕ + sin ϕ tan ϕ )
g
(1 + tan 2 ϕ ) =
2
v02 (1 − sin ϕ )
g cos ϕ
2
(1 + tan ϕ ) .
2
G
58. The position of a particle moving in the xy plane is given by r = ( 2.0cos3.0t,2.0sin3.0t ) , where r is in meters and t is in seconds. (a) Show that this represents circular motion of radius 2.0m centered at the origin. (b) Determine the velocity and acceleration vectors as functions of time. (c) Determine the speed and magnitude of the acceleration. (d) Show that a = v 2 / r . (e) Show that the acceleration vector always points toward the centre of the circle. Solution: (a) From the problem, we have x 2 + y 2 = 4.0 m2 , so this represents circular G G
motion of radius 2.0 m. (b) The velocity reads v = dr dt = ( −6.0sin3.0t,6.0cos3.0t ) , and G G
the acceleration is a = dv dt = ( −18sin3.0t, −18cos3.0t ) . (c) v = vx2 + vy2 = 36 m/s; and a = ax2 + a y2 = 18 m/s 2 . (d) Since r = x 2 + y 2 = 2.0 m, so one gets the relation a = v 2 / r holds. (e) It can be easily to show that the inner product of velocity and G G
acceleration is zero, v ⋅ a = 0 , so they are perpendicular to each other. Since the velocity is always along the tangential of the circle, the acceleration is therefore perpendicular to the velocity and thus points toward the centre of the circle. 66. A boat, whose speed in still water is 2.40 m/s, must cross a 280‐m‐wide river and arrive at a point 120 m upstream from where it starts. To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river's current? G
Solution: If vBS is the velocity of the boat with respect G
to the shore, vBW the velocity of the boat with respect G
to the water, and vWS the velocity of the water with respect to the shore, then G
G
G
vBS = vBW + vWS . Thus we can plot Fig. 3.66. And From Fig. 3.66, we get sin A = sin (π − A ) =
280
280 + 120
2
2
=
7
2
2
.
, sin B = sin ( 3π / 4 − A ) =
( cos A + sin A ) =
2
29 58 From sine theorem, we can get vWS =
vBW
× sin B = 0.97 m/s .
sin A
90. A projectile is launched from ground level to the top of a cliff which is 195m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, find the initial velocity of the projectile. Neglect air resistance. Solution: The horizontal component of the initial velocity is v0 x =
195
m/s = 25.7 m/s. 7.6
And for the vertical component, one can get y = v0 y t −
1 2
gt = 155 m. Thus one can 2
1
⎞
⎛
figure out that v0 y = ⎜ 155 m + gt 2 ⎟ / t = 57.6 m/s. The initial velocity is therefore 2
⎝
⎠
G
v = ( 25.7,57.6 ) m/s, and the magnitude is v = 25.7 2 + 57.62 m/s = 63.1 m/s , and the directional cosine is cos θ =
v0 x
= 0.407 . v0