# 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

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)

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)