Question

The velocity of a particle moving along an axis is given by v(t)=2-t² for t>0, what is the average velocity of the particle from t=1 to t=3?

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​

Answer 1

$\sf\huge\bold{Answer:}$

Given :

• Velocity of particle $\tt\ :⟶v(t) = 2 - t {}^{2}$
• t>0

To find :

• the average velocity of the particle from t=1 to t=3

Solution :

$\fbox{Value of v [when t=1]}$

$\tt\ :⟶v(t) = 2 - t {}^{2}$

$\tt\ :⟶v(1) = 2 - (1){}^{2}$

$\tt\ :⟶v(1) = 2 -1$

$\tt\ :⟶v(1) = 1ms {}^{ - 1}$

$\fbox{Value of v [when t=3]}$

$\tt\ :⟶v(t) = 2 - t {}^{2}$

$\tt\ :⟶v(3) = 2 - (3){}^{2}$

$\tt\ :⟶v(3) = 2 - 9$

$\tt\ :⟶v(3) = - 7ms {}^{ - 1}$

Average velocity :

$\tt\red{ :⟶v _{avg.} = final \: velocity - initial \: velocity}$

$:⟶ \tt\ v _{avg.} = v(3) - v(1)$

• Inserting values of v(3) and v(1)

$\tt\ :⟶v _{avg.} = - 7 - ( 1)$

$\tt\ :⟶v _{avg.} = - 8ms {}^{ - 1}$

Magnitude of average velocity is 8m/s and direction is opposite.[negative sign].