Question

The position of a 2.75×105N training helicopter under test is given by r⃗ =(0.020m/s3)t3i^+(2.2m/s)tj^−(0.060m/s2)t2k^.

Answer 1

Answer:

$16819.57185\hat{i}+3363.91437\hat{j}$

Explanation:

The position vector is

$\vec{r}=0.02t^3\hat{i}+2.2t\hat{j}+0.06t^2\hat{k}$

Differentiating with respect to time

$v=\dfrac{d}{dt}0.02t^3\hat{i}+2.2t\hat{j}+0.06t^2\hat{k}\\\Rightarrow v=0.06t^2\hat{i}+2.2\hat{j}+0.12t\hat{k}$

Differentiating with respect to time

$a=\dfrac{dv}{dt}\\\Rightarrow a=\dfrac{d}{dt}0.06t^2\hat{i}+2.2\hat{j}+0.12t\hat{k}\\\Rightarrow a=0.12t\hat{i}+0.12\hat{k}$

Mass of the helicopter

$m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{2.75\times 10^5}{9.81}\\\Rightarrow m=28032.6197757\ kg$

Net force on the helicopter

$F=ma\\\Rightarrow F=28032.6197757(0.12t\hat{i}+0.12\hat{k})\\\Rightarrow F=3363.91437\hat{i}t+3363.91437\hat{j}$

At t = 5 s

$F=3363.91437\hat{i}\times 5+3363.91437\hat{j}\\\Rightarrow F=16819.57185\hat{i}+3363.91437\hat{j}$

The force is $16819.57185\hat{i}+3363.91437\hat{j}$