Question

if a body travels with speed v (1),v (2),v (3) for equal intervals of time ,find the average speed . ​

Answer 1

Given :

• A body travelled with speeds v₁ , v₂ and v₃ for equal intervals of time

To Find :

• The average speed of the body

Solution :

Let the time interval be "t"

Since ,

$\\ \star \: {\boxed{\purple{\sf{distance = speed \times time}}}}$

Distance covered with speed v₁ in time t is v₁t

Distance covered with speed v₂ in time t is v₂t

Distance covered with speed v₁ in time t is v₃t

Total Time taken in these three intervals is t + t + t = 3t [given condition]

Now , Average speed of a body is nothing but the ratio between total distance travelled and total time taken.If the Mathematically ,

$\\ \star \: {\boxed{\purple{\sf{ v_{avg} = \dfrac{total \: distance}{total \: time \: taken}}}}}$

Now , Calculating Average speed of given body ;

$\\ : \implies \sf v_{avg} \: = \dfrac{v_1t + v_2t + v_3t }{t + t + t}$

Taking t common in numerator ,

$\\ : \implies \sf \: v_{avg} = \dfrac{t(v_1 + v_2 + v_3)}{3t}$

Cancelling t on both numerator and denominator ,

$\\ : \implies{\underline{\boxed{\mathfrak{ \pink{ \: v_{avg} = \dfrac{v_1 + v_2 + v_3}{3} }}}}} \: \bigstar$

$\\ \therefore {\underline{\sf{Hence \: , \: The \: average \: speed \: of \: the \: given \: body \: is \: \bold{ \frac{v_1 + v_2 + v_3}{3} }}}}$

Answer 2

Given:-

If a body travels with speed v (1) , v (2) , v (3) for equal intervals of time

To Prove:-

The average speed?...

Solution:-

To find total distance: -

Using Formula,

$\large \star{ \boxed{ \boxed{ \sf{ \pink{distance = speed \times time}}}}}$

$\large \star{ \boxed{ \boxed{ \sf{distance = v _1 \times t \: = v_1t }}}}$

$\large \star{ \boxed{ \boxed{ \sf{distance = v _2 \times t \: = v_2t }}}}$

$\large \star{ \boxed{ \boxed{ \sf{distance = v _3 \times t \: = v_3t }}}}$

Total time taken in three intervals:-

$\large \star{ \boxed{ \boxed{ \sf{ \pink{total \: time = t + t + t = 3t}}}}}$

To find average speed:-

$\large \star{ \boxed{ \boxed{ \sf{ \pink{v_a \: _v \:_g = \frac{total \: distance}{total \: time} }}}}}$

$\large \star{ \boxed{ \boxed{ \sf{v_a \: _v \:_g = \frac{t \: (v_1 + v_2 + v_3)}{3t} }}}}$

$\large \star{ \boxed{ \boxed{ \sf{v_a \: _v \:_g = \frac{ \: v_1 + v_2 + v_3}{3} }}}}$

Hence, the average speed is $\bold{v_a \: _v \:_g = \frac{ \: v_1 + v_2 + v_3}{3}}$