Phys 1001 (Homework 1)
1) Correctly quote the following quantities (10 points):C
a) $2.452 +/- 0.0935$ in
b) $136798 +/- 5598$
c) $549.0153 +/- 34.021$ N
d) $0.01524 +/- 0.01341$ kg
e) $55.672 +/- 3.96 (ms^{-1})$
2) Convert the following quantities to SI units (kg, m, s). Assume 1 mile = 1609 m, 1 ton
1000 kg, 1 inch = 2.54 cm, 1 foot = 0.3048 In, and 1 pound = 0.454 kg. (15 points) – i
a) 5.2 miles per minute.
b) 18.3 tons per square inch
c) 21.0 pounds per year.
d) 139.8 cubic feet per decade.
6) 3.673 x 10‘ grams per week.
3) You construct a pile of stacked wood and measure it be 1.00 +/- 0.05 m long by 1-00 +/- 05 wide by 0.50 +/- 0.04 m high. What is the volume of the pile, With uncertainty? [5 P“)
4) Find S= A + B where A is 50 m at 30° and B is 30 mat 70°. (5 points)
5) A pilot must travel directly to a town 320 miles away in a direction 10° east of south
from her present position. There is a steady 30 mph wind blowing west to east. Her plane’s
cruising speed through calm air is 95 mph. Using vectors, what must be the plane’s
heading? (5 points)
6) Two vectors are A = 3i +4j; and B = i + 7 j . What is the angle between them? (5 points)
7) A bicycle racer is moving at a speed of 14 m/s. To pass another cyclist. the racer speeds up with an acceleration of 1.2 m/s. At the end of 6.0 s, how fast is the racer moving? (.5 points)
8) Harry stands at the edge of the Grand Canyon and dumps a large rock over the edge. A
Exactly 7.0 s later he sees the rock it the bottom. How deep is the canyon here? (8 points)
9) Two cars, A and B, are moving in the same direction. When t = 0, their respective
velocities are 1 m/s and 3 m/s and their respective accelerations are 2 m/s^2 and 1 m/s^2. If car A is 1.5 m ahead of car B at t: 0, calculate when and where they will be side to side.
(12 points)
Answers
1.
a) $2.45 +/- 0.09$
b) $136000 +/- 5000$
c) $540 +/- 30$
d) $0.01 +/- 0.013$
e) $55 +/- 3$
2)
a) $5.2 (miles/minute) =5.2*1609/60 (m/s) =139.447 (m/s)$
b) $18.3 tons/sq. in = 18.3*1000/(0.0254)^2 = 28365056.7 kg/m^2$
c) $210 lb/year = 210*0.454/(3600*24*365) =3.02*10^{-6} kg/s$
d) $139.8 cubic ft/decade = 139.8 *(0.3048)^3/(10*365*24*3600) =1.255*10^{-8} m^3/s$
e) $3.673*10^6 grams/week =3.673*10^3/(7*24*3600) =0.00607 kg/s$
3)
Volume is
$V = x y z = 1.00*1.00*0.50 =0.5 (m^3)$
uncertainty is
$d V = y z*d x +x z*d y + x y*dz = 1*0.5*0.05 +1*0.5*0.05 +1*1*0.04 =0.09 (m^3)$
Volume $V = 0.50 +/- 0.09 (m^3)$
4.
$A_x = A*cos(30) = 50*cos(30) =43.30$
$A_y = A*sin(30) = 50*sin(30) = 25$
$B_x = B*cos(70) =30*cos(70) =10.26$
$B_y = B*sin(70) =30*sin(70) =28.19$
$(A+B)_x = 43.30+10.26 =53.56$
$(A+B)_y = 53.19$
$S = A+B = sqrt{53.56^2 +53.19^2} =75.484$
5.
Direction west-east is x (+i)
Direction South-North is y (+j)
See attached figure.
The angle of the resultant speed R with x is -80 degree (80 deg south of east)
The speed of the wind is $W = W_x = 30$
The speed of plane is V.
$V/sin(80) = W/sin(x)$
$sin(x) = W/V*sin(80) = 30/95* sin(80) =0.31$
$x =18.11 deg$
It means the airplane heading need to be
$80+18.11 =98.11$ degree south of east or
equivalent $(90-98.11) = 8.11$ degree West of south
(in the picture the direction of the V should be downward to the west)
6)
$tan(a) = 4/3$ $a =53.13(deg)$ from +x direction
$tan(b) =7/1$ $b =81.87 (deg)$ from +x direction
angle between vectors A and B is
$x = b -a =81.87-53.13 =28.74$ degree
7.
$V = V_0 +a*t = 14 +1.2*6 = 21.2 (m/s)$
8.
$S = g*t^2/2 = 9.81*7*7/2 =240.345 m$
9.
distance traveled by car A is
$D_1 = S_0 + v(A)*t +a(A)*t^2/2 = 1.5 + 1*t + 2*t^2/2$
distance traveled by car B is
$D_2 = V(B)*t +a(B)*t^2/2 = 3*t +1*t^2/2$
The two distances need to be equal
$1.5 + t + t^2 = 3*t +t^2/2$
$3 +2t +2t^2 =6t +t^2$
$t^2 -4t + 3 =0$
$t_1 = 2 – sqrt{4-3} = 2 – sqrt{1} =1$
$t_2 = 2+ 1= 3$ seconds
The distances from B car position are
$D(1) = 1.5 + 1+ 1 =3.5$ meters (from B car position)
$D(2) = 1.5 +3 + 9 = 13.5$ meters (from B car position)
($t_2$ is negative and has no physical meaning)
