UNIVERSITY OF JOHANNESBURG ! APM 1B SEMESTER TEST 1

UNIVERSITY OF JOHANNESBURG - APM 1B
SEMESTER TEST 1 solutions
21 August 2014
All physical quantities are in SI units. Angles are in radians
QUESTION 1
[13]
r (t) =
t3 ; sin t
v (t) =
3t2 ; cos t
a (t) = ( 6t; sin t)
(a)
r (1) = ( 1; 0:8415)
v (1) = ( 3; 0:5403)
a (1) = ( 6; 0:8415)
(b)
r (1) = 1:3069
v (1) = 3:0483
a (1) = 6:0587
(c)
rb =
(d)
r (1)
= ( 0:7652; 0:6439)
r (1)
b = ( 0:6439; 0:7652)
r_ = v (1) rb = 2:6433
2
r• = a (1) rb + r _ = 5:8126
(e)
b
_ = v (1)
= 1:1616
r
b 2r_ _
• = a (1)
= 1:2503
r
rb = (cos ; sin )
) rb x
b = cos
)
= arccos (( 0:7652; 0:6439) (1; 0)) = arccos ( 0:7652)
= 2:4421 rad (= 139:92 deg)
QUESTION 2
[12]
cos t; 6t; et + 2
Z
v (t) =
a (t) dt = sin t; 3t2 ; et + 2t + A
Z
r (t) =
v (t) dt =
cos t; t3 ; et + t2 + At + B;
a (t) =
where A and B are vectors of constants.
r (0) = ( 1; 0; 1) + B = (0; 0; 0)
) B = (1; 0; 1)
and
r (1) = ( cos 1; 1; e + 1) + A + (1; 0; 1) = (3; 2; 1)
) A = (2 + cos 1; 1; 1 e) = (2:540; 1; 1:718) :
Hence,
v (0) = (0; 0; 1) + (2:540; 1; 1:718)
= (2:540; 1; 0:718) :
____________________________________________
Information provided:
dr
= r_
dt
dv
a =
= v_
dt
v =
v = rb
_r + r _ b
rb = (cos ; sin )
a = r•
r_
v = jvj
2
rb + r• + 2r_ _ b
b = ( sin ; cos )
r=
Z
vdt + C

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