Question

plz solve this numerical
Class 9th

Answer 1

I hope it will help you

Answer 2

$\huge{\pink{\underline{\ulcorner{\star\:Answer\:\star}\urcorner}}}$

Let we consider two cars A and B

$\red{\underline{Car \:A }}$

velocity = 52 km/h

time = 5 sec (when brake applies take 5 sec in stopping)

$\red{\underline{Car \:B }}$

velocity = 3 km/h

time = 3 sec (when brake applies take 10 sec in stopping)

$\red{\underline{Convert \:velocity \:into\: km\:h^{-1}\: to\: \:m\:sec^{-1}}}$

$velocity (1) = \frac{5}{18}\times 52\\\implies 14.4\: m\: sec^{-1}$

$velocity (2) = \frac{5}{18}\times 3\\\implies 0.83\: m\: sec^{-1}$

Distance travelled by A (s) = Area under triangle AOC

$Area\: under \:triangle (AOD) = \frac{1}{2}(AO)(OC)\\\implies \frac{1}{2}(14.4)(5) \\\implies 7.2\times 5\\\implies 36.00 m$

Distance travelled by B (s) = Area under triangle BOD

$Area\: under \:triangle (BOD) = \frac{1}{2}(BO)(OD)\\ \implies\frac{1}{2}(0.83)(10) \\\implies 0.83\times 5\\\implies 4.15 m$

$\red{\underline {Statement \:first\colon}}$

The car A travels 36 m after applying the brakes on the other hand the car B takes 4.1 m distance in stopping or applying brake.

$\red{\underline{Statement\: Second\colon}}$

From the statement first we learnt that the car A travels farther then car B after applying the force

(Note the v-t graph is in attachment please take a Refrence from it)

thanks (have a nice day)