Question

A particle moving with constant acceleration of 2m/s2 due west has an initial velocity of 9m/s due east.Find the distance covered in the fifth second of its motion

Answer 1

u=9m/s
a=2m/s2
S5=u+a(n-1/2)
=9+2(5-1/2)
=9+2*4.5
=9+9=18

hence the distance covered in fifth second is 18 m.

Answer 2

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⟹ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴜ=± 9 ᴍ/ꜱ

⟹ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ @=-2ᴍ/ꜱ 2

⟹ɪɴ ᴛʜɪꜱ ᴘʀᴏʙʟᴇᴍ , ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴅɪʀᴇᴄᴛɪᴏɴ ɪꜱ ᴏᴘᴘᴏꜱɪᴛᴇ ᴛᴏ ᴛʜᴇ ᴠᴇʟᴏᴄɪᴛʏ ᴅɪʀᴇᴄᴛɪᴏɴ.

⟹ ʟᴇᴛ 'ᴛ' ʙᴇ ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ʙʏ ᴛʜᴇ ᴘᴀʀᴛɪᴄʟᴇ ᴛᴏ ʀᴇᴀᴄʜ ᴀ ᴘᴏɪɴᴛ ᴡʜᴇʀᴇ ɪᴛ ᴍᴀᴋᴇꜱ ᴀ ᴛᴜʀɴ ᴀʟᴏɴɢ ᴛʜᴇ ꜱᴛʀᴀɪɢʜᴛ ʟɪɴᴇ.

⟹ ᴡᴇ ʜᴀᴠᴇ , ᴠ = ᴜ ± @ᴛ

0 = 9-2ᴛ

⟹ᴡᴇ ɢᴇᴛ, ᴛ= 4.5 ꜱ

⟹ɴᴏᴡ ʟᴇᴛ ᴜꜱ ꜰɪɴᴅ ᴛʜᴇ ᴅɪꜱᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ 1/2 ꜱᴇᴄᴏɴᴅꜱ ɪ.ᴇ ꜰʀᴏᴍ 4.5ᴛᴏ 5 ꜱᴇᴄᴏɴᴅꜱ.

ʟᴇᴛ ᴜ = 0 ᴀᴛ ᴛ= 4.5 ꜱᴇᴄ.

ᴛʜᴇɴ ᴅɪꜱᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ 1/2 ꜱ.

ꜱ= 1/2 aᴛ²

ꜱ = 1/2 × 2 × [1/2]²

⟹ 1/4 ᴍ

ᴛᴏᴛᴀʟ ᴅɪꜱᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ ꜰɪꜰᴛʜ ꜱᴇᴄᴏɴᴅ ᴏꜰ ɪᴛꜱ ᴍᴏᴛɪᴏɴ ɪꜱ ɢɪᴠᴇɴ ʙʏ

⟹ ꜱ0 = 2ꜱ = 2(1/4) = 1/2 ᴍ.