Question

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Answer 1

Answer on Question # 42891, Physics, Mechanics | Kinematics | Dynamics

Task:

19. The distance $x$ of a particle moving in one dimension under the action of constant force is related to the time $t$ by the relation, $t = \sqrt{x} + 3$ . Find the displacement of the particle its velocity is $6.0 \, \text{m/s}$ :

(a)9.0m

(b)6.0m

(c)4.0m

(d)0.0m

Solution:

$$x = (t - 3) ^ {2} \Rightarrow \frac {d x}{d t} = V = 2 (t - 3) \Rightarrow V = 6. 0 m / s \Rightarrow t = 6. 0 s.$$ $$t = 6. 0 = \sqrt {x} + 3 \Rightarrow x = 9. 0 m.$$

Answer: (a)9.0m

20. Which of the following pairs of vectors are parallel?

(a) $\vec{A} = \hat{i} - 2\hat{j}; \vec{B} = \hat{i} - 5\hat{j}$

(b) $\vec{A} = \hat{i} - 10\hat{j}; \vec{B} = 2\hat{i} - 5\hat{j}$

(c) $\vec{A} = \hat{i} - 5\hat{j}; \vec{B} = \hat{i} - 10\hat{j}$

(d) $\vec{A} = \hat{i} - 5\hat{j}; \vec{B} = 2\hat{i} - 10\hat{j}$

Solution:

if vectors are parallel then $\angle (\vec{A},\vec{B}) = \arccos \frac{\vec{A}\cdot\vec{B}}{|\vec{A}|\cdot|\vec{B}|} = 0.$

(a) $\angle (\vec{A},\vec{B}) = \arccos \frac{(\hat{i} - 2\hat{j})(\hat{i} - 5\hat{j})}{|\hat{i} - 2\hat{j}|\cdot|\hat{i} - 5\hat{j}|} = \frac{52}{\sqrt{101\cdot 29}}\neq 0$

(b) $\angle (\vec{A},\vec{B}) = \arccos \frac{(\hat{i} - 10\hat{j})(2\hat{i} - 5\hat{j})}{|\hat{i} - 10\hat{j}|\cdot|2\hat{i} - 5\hat{j}|} = \frac{51}{\sqrt{101\cdot 26}}\neq 0$

(c) $\angle (\vec{A},\vec{B}) = \arccos \frac{(\hat{i} - 5\hat{j})(\hat{i} - 10\hat{j})}{|\hat{i} - 5\hat{j}|\cdot|\hat{i} - 10\hat{j}|} = \frac{52}{\sqrt{104\cdot 26}}\neq 0$

(d) $\angle (\vec{A},\vec{B}) = \arccos \frac{(\hat{i} - 5\hat{j})(2\hat{i} - 10\hat{j})}{|\hat{i} - 5\hat{j}|\cdot|2\hat{i} - 10\hat{j}|} = \frac{52}{\sqrt{104\cdot 26}} = 0$

Answer: (d)

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