Motion in a Straight Line - Result Question 15
15. A particle has initial velocity $(3 \hat{i}+4 \hat{j})$ and has acceleration $(0.4 \hat{i}+0.3 \hat{j})$. It’s speed after 10 $s$ is:
[2010]
(a) 7 units
(b) $7 \sqrt{2}$ units
(c) 8.5 units
(d) 10 units
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Answer:
Correct Answer: 15. (b)
Solution:
- (b) Given, $\vec{u}=3 \hat{i}+4 \hat{j}$ and $\vec{a}=0.4 \hat{i}+0.3 \hat{j}$
$\Rightarrow u_x=3$ units, $u_y=4$ units
$\Rightarrow a_x=0.4$ units, $a_y=0.3$ units
Along $x$-axis,
$\therefore v_x=u_x+a_x \times 10=3+4=7$ units
Along $y$-axis,
and $v_y=4+0.3 \times 10=4+3=7$ units
Net final velocity $\therefore \quad v=\sqrt{v_x^{2}+v_y^{2}}=7 \sqrt{2}$ units
$ \vec{v} _i=3 \hat{i}+4 \hat{j}$ and $\vec{a}=0.4 \hat{i}+0.3 \hat{j}$
time, $t=10 sec$.
Final velocity $ \vec{v} _f$ after time $t=10 sec, \vec{v} _f= \vec{v} _i+\vec{a} t$
$ \vec{v} _f=(3 \hat{i}+4 \hat{j})+(0.4 \hat{i}+0.3 \hat{j})(10)=7 \hat{i}+7 \hat{j}$
The particle speeds up i.e., the speed of the particle increases when the angle between $\vec{a}$ and $\vec{v}$ lies between $0^{\circ}+90^{\circ}$. The particle speeds down i.e., the speed of the particle decreases when the angle between $\vec{a}$ and $\vec{v}$ lies between $+90^{\circ}$ and $180^{\circ}$.
