Question

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

Answer 1

Answer:

See below

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

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

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$\large \tt Solution \: :$

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 ]

$\qquad \tt \rightarrow \:v = \dfrac{s}{t}$

$\qquad \tt \rightarrow \:v = \dfrac{4 \hat i + 3 \hat j}{2}$

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

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 ❞