Physics, asked by choudhryhello, 15 days ago

A body is travelling with a velocity of 5m/s toward the east and changes to 5m/s toward the north in 10 seconds. Find acceleration of the body .

Answers

Answered by YourHelperAdi
3

Given :

  • Velocity while going east = 5m/s
  • Velocity while going north = 5m/s
  • Time taken = 10 seconds

To Find :

  • The acceleration of the body

Always Remember:

Here, you may be confused about here that , the acceleration may be zero, as the Velocity are same.

But : Remember that, Velocity is a vector quantity, Hence, it requires the direction also.

Hence, the acceleration also depends upon direction.

\displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ \overrightarrow{v_1} -  \overrightarrow{v _2}}{t} |

Formula To Be Applied :

We will Apply the formula of subtraction of vectors :

  \displaystyle \rm  r =  \sqrt{ {a}^{2} +  {b}^{2}   - 2ab (cos \theta)}

Where,

r = magnitude of Resultant vector

a= magnitude of first vector

b = magnitude of second vector

theta = Angle between (a and b)

Solution :

As we know, acceleration = (change of Velocity upon time)

Given, Change in Velocity = Velocity changes from east to north.

Time = 10s

\displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ \overrightarrow{v_1} -  \overrightarrow{v _2}}{t} |

 \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ \overrightarrow{r}}{t} |

Now, we will calculate vector 'r' , by using the formula as stated .

{ \implies \displaystyle \rm   | \overrightarrow{r}|  =  | \sqrt{( v_1   {)}^{2} + (v_2  {)}^{2}  - 2(v_1 )(v_2)(cos \theta)}| }

{ \implies \displaystyle \rm   | \overrightarrow{r}| =  \sqrt{ {5}^{2}  +  {5}^{2}  + 2(5)(5)(cos 9 {0}^{ \circ}) } }

 {\implies \displaystyle \rm   | \overrightarrow{r}| =  \sqrt{25 + 25 + 2(25)(0)} }

 \implies\displaystyle \rm   | \overrightarrow{r}| =  \sqrt{25 + 25 - 0}

 \implies \displaystyle \rm   | \overrightarrow{r}| =  \sqrt{50}

 \implies \displaystyle \rm   | \overrightarrow{r}| =  \sqrt{5 \times 5 \times 2}

 \implies \displaystyle \rm   | \overrightarrow{r}| = 5 \sqrt{2}

So, as we got the resultant, now using this we will find the acceleration.

 \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ \overrightarrow{r}}{t} |

 \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ 5 \sqrt{2} }{10} |

 \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ \cancel{5} \sqrt{2} }{ \cancel{10}} |

  \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{  \sqrt{2} }{2} |

  \blue{ \underline{ \boxed{ \implies \displaystyle \rm   | \overrightarrow{a}|  =  | \frac{ 1}{ \sqrt{2} } | }}}

Hence, acceleration = 1/2 m/s North East

Answered by choudhrybye
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