Question

please answer 12th and 13th Question​

Answer 1

⟶ 12th AnSwer :-

A bus is moving with a speed of 10m/son a straight road. A scooterist wishes to overtake the bus in 10 seconds. if the bus is at a distance of 1km from the scooterist with what speed should the scooterist chase the bus ?

☑ (1) 20m/s

(2) 40m/s

(3) 25m/s

(4) 10m/s

Option (1) 20m/s is right

Explaination :-

The relative velocity of scooter w.r.t bus,

$\longrightarrow \sf \bar { v_{sb} } = \bar { v_{s} } - \bar { v_{b} } = \bar { v_{s} } - 10 \: \: \: \: \: \: ........(1)$

$\longrightarrow \sf relative \: velocity = \frac{relative \: displacement}{time}$

$\longrightarrow \sf \bar { v_{s} } - 10 = \frac{1000}{100} = 10$

$\longrightarrow \sf \bar { v_{s} } = 10 + 10 = 20 {ms}^{ - 1}$

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⟶ 13th AnSwer :-

A particle starts its Motion from rest under the action of constant force .distance cover in first 10 s is $\sf s_1$and that covered in the first 20s is $\sf s_2$ , then

(1) $\sf s_2$= 2 $\sf s_1$

(2) $\sf s_2$= 3 $\sf s_1$

☑ (3) $\sf s_2$= 4 $\sf s_1$

(4) $\sf s_2$= $\sf s_1$

Option (3) $\sf s_2$= 4 $\sf s_1$ is right.

Explaination :-

lf the particle is moving in a straight line under the action of constant force or under constant acceleration a using

$\longrightarrow \sf s = ut + \frac{1}{2} {at}^{2}$

Since the body starts from the rest u = 0

$\therefore \sf s = \frac{1}{2} {at}^{2}$

Now,

$\longrightarrow \sf s_{1} = \frac{1}{2} a(10) ^{2} \: \: \: \: \: .........(1) \\ \\ \longrightarrow \sf and \: s_{2} = \frac{1}{2} a(20)^{2} \: \: \: \: ........(2) \: \: \: \: \: \: \: \: \: \:$

Dividing eq (1) and (2) we get,

$\longrightarrow \sf \frac{ s_{1} }{s_{2}} = \frac{ \cancel \frac{1}{2} a {(10)}^{2} }{ { \cancel \frac{1}{2}a (20)}^{2} } \\ \\ \longrightarrow \sf\frac{ s_{1} }{s_{2}} = \frac{ {(10)}^{2} }{ {(20)}^{2} } = \frac{100}{400} = \frac{1}{4} \\ \\ \longrightarrow \sf \frac{ s_{1} }{s_{2}} = \frac{1}{4} \\ \\ \longrightarrow \sf { s_{2} } = 4{s_{1}}$