Question

Bob travel 4m to the east and 3m to the north. what is the velocity of Bob in 2sec?​

Answer 1

See below

Step-by-step explanation:

He travels a total distance of 4 + 3 = 7 m in 2 seconds

(Note: this is the actual distance he traveled NOT the displacement from his origin)

velocity = 7m/2s = 7/2 m/s

From starting point he has traveled sqrt ( 3^2+4^2) = 5m in 2 seconds

so his RESULTANT VELOCITY :

5/2 m/s at angle tan^-1( 3/4) = 36.86° North of east

Answer 2

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

Bob's Velocity :

`\qquad \tt \rightarrow \:v = 2 \hat i + \cfrac{3 \hat j}{2}`

and

`\qquad \tt \rightarrow \:mag(v) = { 2.5 }^{} \: \: m/s`

____________________________________

`\large \tt Solution \: :`

Let the unit vector towards north be
`\sf \hat j` and unit vector towards east be
`\sf \hat i`

According to statement :

`\qquad \tt \rightarrow \:s = 4 \hat i + 3 \hat j`

[ s represents displacement ]

`\qquad \tt \rightarrow \:v = (s)/(t)`

`\qquad \tt \rightarrow \:v = (4 \hat i + 3 \hat j)/(2)`

`\qquad \tt \rightarrow \:v = 2 \hat i + \cfrac{3 \hat j}{2}`

If magnitude of velocity is to be asked :

`\qquad \tt \rightarrow \:mag(v) = \sqrt{(2) {}^(2) + { \bigg( \cfrac{3}{2} \bigg) }^(2) }`

`\qquad \tt \rightarrow \:mag(v) = \sqrt{4 + { \cfrac{9}{4} }^{} }`

`\qquad \tt \rightarrow \:mag(v) = \sqrt{ { \cfrac{16 + 9}{4} }^{} }`

`\qquad \tt \rightarrow \:mag(v) = \sqrt{ { \cfrac{25}{4} }^{} }`

`\qquad \tt \rightarrow \:mag(v) = { \cfrac{5}{2} }^{} \: \: m/s`

`\qquad \tt \rightarrow \:mag(v) = { 2.5 }^{} \: \: m/s`

Answered by : ❝ AǫᴜᴀWɪᴢ ❞