Question

25. Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (u) and the braking capacity, or deceleration,. A car travelling at speed 72km/hr suddenly applies the brake with the deceleration of 5m/s2 . Find the stopping distance of the car

Answer 1

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Answer 2

$\huge \underline \mathrm \purple{Question↣}$

sᴛᴏᴘᴘɪɴɢ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴠᴇʜɪᴄʟᴇs : ᴡʜᴇɴ ʙʀᴀᴋᴇs ᴀʀᴇ ᴀᴘᴘʟɪᴇᴅ ᴛᴏ ᴀ ᴍᴏᴠɪɴɢ ᴠᴇʜɪᴄʟᴇ, ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ɪᴛ ᴛʀᴀᴠᴇʟs ʙᴇғᴏʀᴇ sᴛᴏᴘᴘɪɴɢ ɪs ᴄᴀʟʟᴇᴅ sᴛᴏᴘᴘɪɴɢ ᴅɪsᴛᴀɴᴄᴇ. ɪᴛ ɪs ᴀɴ ɪᴍᴘᴏʀᴛᴀɴᴛ ғᴀᴄᴛᴏʀ ғᴏʀ ʀᴏᴀᴅ sᴀғᴇᴛʏ ᴀɴᴅ ᴅᴇᴘᴇɴᴅs ᴏɴ ᴛʜᴇ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ (ᴜ) ᴀɴᴅ ᴛʜᴇ ʙʀᴀᴋɪɴɢ ᴄᴀᴘᴀᴄɪᴛʏ, ᴏʀ ᴅᴇᴄᴇʟᴇʀᴀᴛɪᴏɴ,. ᴀ ᴄᴀʀ ᴛʀᴀᴠᴇʟʟɪɴɢ ᴀᴛ sᴘᴇᴇᴅ 72ᴋᴍ/ʜʀ sᴜᴅᴅᴇɴʟʏ ᴀᴘᴘʟɪᴇs ᴛʜᴇ ʙʀᴀᴋᴇ ᴡɪᴛʜ ᴛʜᴇ ᴅᴇᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴏғ 5ᴍ/s² . ғɪɴᴅ ᴛʜᴇ sᴛᴏᴘᴘɪɴɢ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄᴀʀ.

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Here $\bold{u \: =72 km/hr = \frac {72×1000}{3600} m/s \: =20 m/s }$

$\bold{v\: =0}$

$\bold{a = -5 m/s^2}$

Now using the relation ;

$\boxed {\bold{v^2=u^2+2as}}$

$\implies \bold {0=(20)2+2×(−5)×s}$

$\bold{\red{\implies s=40m}}$

ʜᴇɴᴄᴇ , sᴛᴏᴘᴘɪɴɢ ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄᴀʀ = 40ᴍ