Calculus Solutions EX#2.1

 

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Q.01
Solution:
$$\quad t=10s$$ We shall find the instantaneous velocity  when $t=10s$
From the graph, we see that
at $t_1=5 s, \quad s_1=10 m$
at $t_2=15 s, \quad s_2=50 m$
Instantaneous velocity at $t=10s$ is equal to the slope of the tangent line at $t=10.$
$$
m_{tan}=\frac{s_2-s_1}{t_2-t_1}$$ $$
m_{tan}=\frac{50-10}{15-5}$$ $$
m_{tan}=\frac{40}{10}=4 m/s
$$
Q.02
Solution:
We shall find instantaneous velocity at $t=4 s$ and $t=8 s$.
From the graph, we see that
at $t_1=2 s \quad s_1=0$
at $t_2=10 s, \quad S_2 \approx 90 m.}$
Instantaneous velocity at $t=4s$ is equal to the slope of the tangent line at $t=4.$
$$m_{tan }=\frac{s_2-s_1}{t_2-t_1}$$
$$m_{tan }=\frac{90-0}{10-2}$$
$$m_{tan }=\frac{90}{8}=\frac{45}{4} \mathrm{~m} / \mathrm{s} . $$ From the graph we have
at $t_1=4s \quad s_1=0$
at $t=10s , \quad s_2=140 \mathrm{~m}$
Instantaneous velocity at $t=8s$ is equal to the slope of the tangent line at $t=8.$
$$\quad m_{tan }=\frac{s_2-s_1}{t_2-t_1}$$
$$\quad m_{tan }=\frac{140-0}{10-4}=\frac{140}{6}=\frac{70}{3} \mathrm{~m} / \mathrm{s}$$
$$\quad m _{ tan }= \frac{70}{3} \mathrm{~m} / \mathrm{s}$$
Q.03 The accompanying figure shows the position versus time curve… (a) The average velocity over the interval $0 \leq t \leq 3.$ (b) The values of $t$ at which the instantaneous velocity is zero.
Solution:
$(a)$ The average velocity over the interval $0 \leq t \leq 3$
From the graph, we have
at $t_1=0, \quad s_1=10 \mathrm{cm}$
at $t_2 =3 \mathrm{~s}, \quad s_2=10 \mathrm{~cm} $
$$v_{\text {ave }} & =\frac{s_2-s_1}{t_2-t_1}$$
$$v_{\text {ave }}=\frac{10-10}{3-0}=0 \mathrm{~cm} / \mathrm{s}$$ (b) The values of $t$ at which the instantaneous velocity is zero.
We know that the points where the tangent line is horizontal are the points at which the slope of the tangent line is zero. ie the points where the instantaneous velocity is zero.
From the graph, we have that $$ \text { at } t=0, t=2, t \quad t=4 \text { and } t=8 \text {, }$$ the instantaneous velocity is zero( the tangent lines are horizontal at these points).
$(c)$ The values of $t$ at which the instantaneous velocity $v$ is either maximum or minimum. From the graph we have at $t=1$, the slope is positive and the instantaneous velocity is maximum. At $t=3$, the slope is negative and the instantaneous velocity is minimum.
$(d) $ $$ \begin{aligned} \text {The instantaneous velocity when }$t=3s.$ \end{aligned} $$ At $$t_1=2 \mathrm{~s}, \quad s_1=15 \mathrm{~cm} $$ $$t_2=4 \mathrm{~s}, \quad s_2=5 \mathrm{~cm} $$
$$\begin{aligned} \text { Instantaneous velocity } \approx \frac{s_2-s_1}{t_2-t_1}=\frac{5-15}{4-2}=\frac{-10}{2}=-5 \mathrm{~cm} / \mathrm{s} \end{aligned} $$
Q.04 The accompanying figure shows the position version time curve…
Solution:
$(a)$ Since the slope of the tangent line is decreasing with increasing times the instantaneous velocity is decreasing.
$(b)$ the instantaneous velocity is increasing because the slope of the tangent line is increasing with time.
$(c)$ Since the slope of the tangent line increases with time the instantaneous velocity increases.
$(d)$ Since the slope of the tangent line decreases with increasing time, therefore the instantaneous velocity is decreasing.