Question

if a car covers ⅔ of distance with 60km/hr & remaining with 70km/hr find avg speed​

Answer 1

Answer:

Here, car is travelling different distances at different speeds. So Average speed is not simple sum of speed .

If you care about unit speed unit is km/hr I.e. km is unit of distance and hr is unit of time. So formula of speed will be

speed = distance/time implies

distance = speed *time and hence

time = distance /speed.

We calculate separate time first.

Finally total distance /total time gives right answer.

Answer 2

Answer:

Average speed = 63 km/h

Explanation:

In order to calculate the average speed of the car, we'll be using the formula of average speed. That is,

$\twoheadrightarrow \quad \sf {Speed_{(avg)} = \dfrac{Total \; Distance}{Total \; Time} }$

Let us assume the total distance covered as x.

⇒ Total distance = x

Now, we need to calculate the total time taken. In order to find it so, we need to calculate time taken in both cases.

According to the question,

• ⅔ of distance is covered with 60 km/h

And,

• Remaining distance is covered with 70 km/h

In first case :

⇒ Distance travelled $\sf (S_1)$ = ⅔x

⇒ Speed $\sf (V_1)$ = 60 km/h

So,

$\implies \quad \sf {Time= \dfrac{Distance}{Speed} }$

$\implies \quad \sf {T_1= \dfrac{S_1}{V_1} }$

$\implies \quad \sf {T_1= \dfrac{\cfrac{2}{3}x}{60} \; h }$

$\implies \quad \sf {T_1= \dfrac{2}{3}x \times \dfrac{1}{60} \; h }$

$\implies \quad \sf {T_1= \dfrac{1}{3}x \times \dfrac{1}{30} \; h }$

$\implies \quad \sf {T_1= \dfrac{1}{90}x \; h}$

In second case :

Now, in the second case, the remaining distance is covered with the speed of 70 km/h. So, remaining distance is the distance travelled in this case.

$\implies \sf {Distance_{(Remaining)} = Total \; distance - Distance_{(First \; case)} }$

$\implies \sf {Distance_{(Remaining)} = x - \dfrac{2}{3}x }$

$\implies \sf {Distance_{(Remaining)} = \dfrac{3x - 2x}{3} }$

$\implies \sf {Distance_{(Remaining)} = \dfrac{x}{3} }$

So,

⇒ Distance travelled $\sf (S_2)$ = ⅓x

⇒ Speed $\sf (V_2)$ = 70 km/h

So,

$\implies \quad \sf {Time= \dfrac{Distance}{Speed} }$

$\implies \quad \sf {T_2= \dfrac{S_2}{V_2} }$

$\implies \quad \sf {T_2= \dfrac{\cfrac{1}{3}x}{70} \; h }$

$\implies \quad \sf {T_2= \dfrac{1}{3}x \times \dfrac{1}{70} \; h }$

$\implies \quad \sf {T_2= \dfrac{1}{210}x \; h }$

Therefore, total time will be given by,

$\implies \sf {Total \; Time = T_1 + T_2 }$

$\implies \sf {Total \; Time = \Bigg \{ \dfrac{1}{90}x + \dfrac{1}{210}x\Bigg \} \; h}$

$\implies \sf {Total \; Time = \Bigg \{ \dfrac{7x + 3x}{630} \Bigg \} \; h }$

$\implies \sf {Total \; Time = \dfrac{10x}{630} \; h }$

$\implies \sf {Total \; Time = \dfrac{1x}{63} \; h }$

Therefore, total time taken by the car is 1x/63 hours.

Now, substituting values in the formula of average speed to find the average speed.

$\twoheadrightarrow \quad \sf {Speed_{(avg)} =\left \{ \dfrac{x}{\cfrac{1x}{63}} \right \} \; km/h}$

$\twoheadrightarrow \quad \sf {Speed_{(avg)} = \Bigg \{ x \times \dfrac{63}{1x}\Bigg \} \; km/h}$

$\twoheadrightarrow \quad \sf {Speed_{(avg)} = \Bigg \{ \dfrac{63}{1}\Bigg \}\; km/h }$

$\twoheadrightarrow \quad \bf\underline {Speed_{(avg)} = 63 \; km/h }$

Therefore, average speed of the car is 63 km/h.