clear all; clc; close all
fprintf('Prob # 8 \n\n');
fprintf('acceleration of particle is a=dv/dt=(2*t-1) (m/s^2) \n\n')
fprintf('velocity of particle is v=t^2-t+c1 (m/s) \n\n')
fprintf('at t=0, v=2, therefore c1=2 \n\n')
t=6;
c1=2;
v=t^2-t+c1;
fprintf('v = %g (m/s) \n\n', double(v))
fprintf('answer "c" \n\n')
fprintf('Prob # 9.1 \n\n');
fprintf('tangential component of acceleration at=a*cos(60) (ft/s^2) \n\n');
a=70;
at=a*cosd(60);
fprintf('rate of increase in the plane speed = at = %g (m/s^2) \n\n', double(at))
fprintf('answer "c" \n\n')
fprintf('Prob # 9.2 \n\n');
fprintf('normal component of acceleration an=a*sin(60)=v^2/r (ft/s^2) \n\n');
an=a*sind(60);
fprintf('r=v^2/an (ft) \n\n');
v=400;
r=v^2/an;
fprintf('radius of curvature= r = %g (m) \n\n', double(r))
fprintf('answer "d" \n\n')
fprintf('Prob # 10.1 \n\n');
fprintf('v=dr er+r*dtheta etheta(m/s) \n\n');
fprintf('dtheta=3 (rad/s), ddtheta=0, r=0.4*theta \n\n');
fprintf('therefore dr=0.4*dtheta \n\n');
fprintf('vr=dr \n\n');
dtheta=3;
dr=0.4*dtheta;
fprintf('vr= %g (m/s) \n\n',double(dr));
fprintf('answer "a" \n\n')
fprintf('Prob # 10.2 \n\n');
fprintf('vtheta=r*dtheta \n\n');
theta=pi/3;
r=0.4*theta;
vtheta=r*dtheta;
fprintf('vtheta= %g (m/s) \n\n',double(vtheta));
fprintf('answer "d" \n\n')
fprintf('Prob # 11.1 \n\n');
vA=[0 700];
vB=[0 600];
aA=[0 50];
aB=[600^2/400 -100];
fprintf('vB=vA+vB/A \n\n');
fprintf('vB/A=vB-vA \n\n');
vBA=vB-vA;
fprintf('vBA= %g i + %g j (m/s)\n\n',double(vBA(1)),double(vBA(2)));
fprintf('answer "a" \n\n')
fprintf('Prob # 11.2 \n\n');
fprintf('aB=aA+aB/A \n\n');
fprintf('aB/A=aB-aA \n\n');
aBA=aB-aA;
fprintf('aBA= %g i + %g j (m/s)\n\n',double(aBA(1)),double(aBA(2)));
fprintf('answer "d" \n\n')
fprintf('Prob # 12 \n\n');
fprintf('answer "c" \n\n');
fprintf('Prob # 13 \n\n');
fprintf('v2^2=v1^2+2*a*s, therefore a=(v2^2-v1^2)/(2*s) \n\n');
v1=100e3/3600;
v2=0;
s=50;
a=(v2^2-v1^2)/(2*s);
fprintf('a= %g (m/s^2) \n\n',double(a));
fprintf('by Newtons second law sum(F)=m*a \n\n');
m=1500;
F=m*a;
fprintf('total force needed to stop vehicle F= %g (N)\n\n',double(-F));
fprintf('force is applied equally by each brake, therefore force by each brake = F/4 \n\n');
Findiv=-F/4;
fprintf('individual force by each brake = %g (N) \n\n',double(Findiv));
fprintf('answer "b" \n\n');
fprintf('Prob # 14 \n\n');
fprintf('TA+UA->B=TB , work-energy principle \n\n');
fprintf('0.5*m*vA^2-m*g*h=0.5*m*vB^2, dividing both sides by "m" and rearranging we get \n\n');
fprintf('vB^2=vA^2-2*g*h \n\n');
vA=5;
h=0.8;
g=9.81;
vB=sqrt(vA^2-2*g*h);
fprintf('vB= sqrt(vA^2-2*g*h)= %g (m/s)\n\n',double(vB));
fprintf('answer "c" \n\n');
Prob # 8
acceleration of particle is a=dv/dt=(2*t-1) (m/s^2)
velocity of particle is v=t^2-t+c1 (m/s)
at t=0, v=2, therefore c1=2
v = 32 (m/s)
answer "c"
Prob # 9.1
tangential component of acceleration at=a*cos(60) (ft/s^2)
rate of increase in the plane speed = at = 35 (m/s^2)
answer "c"
Prob # 9.2
normal component of acceleration an=a*sin(60)=v^2/r (ft/s^2)
r=v^2/an (ft)
radius of curvature= r = 2639.32 (m)
answer "d"
Prob # 10.1
v=dr er+r*dtheta etheta(m/s)
dtheta=3 (rad/s), ddtheta=0, r=0.4*theta
therefore dr=0.4*dtheta
vr=dr
vr= 1.2 (m/s)
answer "a"
Prob # 10.2
vtheta=r*dtheta
vtheta= 1.25664 (m/s)
answer "d"
Prob # 11.1
vB=vA+vB/A
vB/A=vB-vA
vBA= 0 i + -100 j (m/s)
answer "a"
Prob # 11.2
aB=aA+aB/A
aB/A=aB-aA
aBA= 900 i + -150 j (m/s)
answer "d"
Prob # 12
answer "c"
Prob # 13
v2^2=v1^2+2*a*s, therefore a=(v2^2-v1^2)/(2*s)
a= -7.71605 (m/s^2)
by Newtons second law sum(F)=m*a
total force needed to stop vehicle F= 11574.1 (N)
force is applied equally by each brake, therefore force by each brake = F/4
individual force by each brake = 2893.52 (N)
answer "b"
Prob # 14
TA+UA->B=TB , work-energy principle
0.5*m*vA^2-m*g*h=0.5*m*vB^2, dividing both sides by "m" and rearranging we get
vB^2=vA^2-2*g*h
vB= sqrt(vA^2-2*g*h)= 3.05025 (m/s)
answer "c"