diff options
Diffstat (limited to '1205/CH11')
-rw-r--r-- | 1205/CH11/EX11.1/S_11_1.sce | 32 | ||||
-rw-r--r-- | 1205/CH11/EX11.2/S_11_2.sce | 30 | ||||
-rw-r--r-- | 1205/CH11/EX11.3/S_11_3.sce | 3 |
3 files changed, 65 insertions, 0 deletions
diff --git a/1205/CH11/EX11.1/S_11_1.sce b/1205/CH11/EX11.1/S_11_1.sce new file mode 100644 index 000000000..882dce009 --- /dev/null +++ b/1205/CH11/EX11.1/S_11_1.sce @@ -0,0 +1,32 @@ +clc;
+function x=displ(t)
+ x=t^3-6*t^2-15*t+40;
+ funcprot(0);
+endfunction
+
+function v=vel(t)
+ v=3*t^2-12*t-15;
+ funcprot(0);
+endfunction
+
+function a=acl(t)
+ a=6*t-12;
+ funcprot(0);
+endfunction
+
+//a0 Time at which V=0
+V=poly([-15,-12,3],'x','coeff');// Velocity polynomial
+[t]=roots(V);// Solution for polynomial
+printf("When V=0 time should be %.0f s or %.0f s \n",t(1),t(2));
+printf("From this only root %.0f s is valid\n",t(1));
+
+//Position and distance traveled when v=0
+x5=displ(t(1));//m, posiyion when V=0
+x0=displ(0);//m, position when t=0
+printf("Position at t=5 s is %.0f m \n and displacement from t=0 to V=0 s is %.0f m \n",x5,x5-x0);
+printf("Negative sign shows that displacement is in negative direction \n")
+printf("Acceleration when v=0 is %.0f m/s^2 \n",acl(t(1)));
+// Diaplacement upto 5 is negative and after that it is positive so we compute it seperately
+d45=displ(5)-displ(4);//m, distance travelled from 4 to 5
+d56=displ(6)-displ(5);//m, displacement from 5 to 6
+printf("Total distance travelled is %.0f m \n",d56-d45);
diff --git a/1205/CH11/EX11.2/S_11_2.sce b/1205/CH11/EX11.2/S_11_2.sce new file mode 100644 index 000000000..69255df33 --- /dev/null +++ b/1205/CH11/EX11.2/S_11_2.sce @@ -0,0 +1,30 @@ +clc;
+a=-9.81;//m/s^2, Acceleration
+// By theoritical work we get following functions
+function y=displ(t)
+ y=20+10*t-4.905*t^2;//m
+ funcprot(0);
+endfunction
+
+function v=vel(t)
+ v=10-9.81*t;//m/s
+ funcprot(0);
+endfunction
+
+//At highest elevation v=0,
+
+t=10/9.81;//s,
+// Putting it in displacement function
+
+printf("When V=0 time should be %.3f s \n",t);
+printf("Highest elevation %.1f m\n",displ(t));
+
+//Ball hits the ground,then y=0
+
+
+Y=poly([20,10,-4.905],'x','coeff');// Velocity polynomial
+[t]=roots(Y);// Solution for polynomial
+//Here only positive value is valid
+t=t(1);
+printf("Time to reach ground is %.2f s\n",t);
+printf("Hitting velocity is %.1f m/s\n",vel(t));
diff --git a/1205/CH11/EX11.3/S_11_3.sce b/1205/CH11/EX11.3/S_11_3.sce new file mode 100644 index 000000000..74eb56bac --- /dev/null +++ b/1205/CH11/EX11.3/S_11_3.sce @@ -0,0 +1,3 @@ +
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");
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