Question

please solve the above questions​

Answer 1

Explanation:

$\bold{Velocity \: of car \: A (V_A) = 36 \: km {h}^{ - 1} }$

$\scriptsize \bold{ = \frac{36000}{60 \times 60} } \: m/s \: ( \because \: 1km = 1000m \: and \: 1h = 60 \times 60sec.)$

$\bold{ = \frac{ \cancel{36000}}{ \cancel{3600} }}$

$\bold{ = 10 \:m/s }$

$\bold{Velocity \: of \: car B (V_B) = 54 \: km \: {h}^{ - 1} }$

$\bold{ = \frac{54000}{60 \times 60} } \: m/s$

$\bold{ = \frac{ \cancel{54000}}{ \cancel{3600}} }$

$\bold{ = 15 \: m/s}$

$\scriptsize\bold{Velocity \: of \: car \: C (V_C) = 54 \: km/h\: (Speed \: of \: car \: B = speed \: of \: car \: C)}$

$\bold{ = \frac{54000}{60 \times 60} } \: m/s$

$\bold{ = \frac{ \cancel{54000}}{ \cancel{3600}} }$

$\bold{ = 15 \: m/s}$

Relative velocity of car C with respect to car A,

$V_{CA} = V_C -(-V_A)$

$= 15 - ( - 10)$

$= 15 + 10$

$= 25 \: m/s$

$\scriptsize \bold{Time(t) \: taken \: by \: car \: C \: to \: overtake \: car \: A = \frac{1000}{25} }$

$= 40 \: sec.$

So, B have to cover distance of (1000m+400m)

i.e. 1400m to take over a A before C does. in 40 sec.

Now putting in formula :-

$\bold{s = ut + \frac{1}{2} a {t}^{2} }$

$\bold{1400 = 15 \times 40 + \frac{1}{2} \times a \times {40}^{2} }$

$\bold{1400 = 600 + \frac{1}{ \cancel2} \times \cancel{1600}a }$

$\bold{1400 - 600 = 800a}$

$\bold{800 = 800a}$

$\bold{a = \frac{800}{800} }$

$\bold{a = 1 \: m/s²}$

See image for 2nd question

Answer 2

Answer:

Answer:

sinα - αcosα

Step-by-step explanation:

Given that,

\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|lim

x→α

∣

x−α

xsinα−αsinx

∣

Let x - α = h

Thus when x ⇒ α , h ⇒ 0

\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|lim

h→0

∣

h

(h+α)sinα−αsin(h+α)

∣

Now we will evaluate left hand limit and right and limit separately

For LHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

−h

αsinα[1−cos(−h)]

+sinα−

−h

αsin(−h)cosα

=

h→0

lim

−h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

−h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

For RHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

h

αsinα[1−cosh]

+sinα−

h

αsinhcosα

=

h→0

lim

h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

Thus limit is (sinα - αcosα).