Question

if I drive from Oxford to Cambridge at 40 km per hour and then from Cambridge to Oxford at 60 km per hour what is my average speed for the whole journey​

Answer 1

Answer:

48 km/hr is the average speed for the entire trip.

Explanation:

First, let's see how average speed is calculated.

$\mathrm{V_{avg}=\frac{D}{T}}$       (1)

Where,

$\mathrm{V_{avg}}$=average speed

D=total distance traveled

T=total time taken

Now, let's suppose that "d" is the one-way distance. Then,

CASE I: From Oxford to Cambridge

$\mathrm{t_1=\frac{d}{40} }$         (2)

CASE II: From Cambridge to Oxford

$\mathrm{t_2=\frac{d}{60} }$         (3)

So,

Total distance(D)=2d        (4)

Total time(T)=t₁+t₂             (5)

By placing equations (4) and (5) in equation (1) we get:

$\mathrm{V_{avg}=\frac{2d}{t_1+t_2}}$      (6)

Equation (6) is obtained by using equations (2) and (3).

$\mathrm{V_{avg}=\frac{2d}{\frac{d}{40} +\frac{d}{60} }}$

$\mathrm{V_{avg}=\frac{2d}{\frac{5d}{120} }}$

$\mathrm{V_{avg}=48\ km/hr}$

So, 48 km/hr is the average speed for the entire trip.

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Answer 2

Answer: The answer is 48 Km/hr

Average Speed: The sum of all of the distances that a thing travels in a given amount of time is the average speed of that object. The speed that is considered to be the norm is a scalar quantity. The magnitude is how it is portrayed, and it does not have any particular orientation.  

Explanation:

Step 1: Given Data:

Speed from Oxford to Cambridge, $v_1 = 40 km/hr$

Speed from Cambridge to oxford, $v_2 = 60km/hr$

Let the distance between them be d.

Step 2: Find the time taken for travel between these places

Time taken from Oxford to Cambridge, $t_1 = \frac{d}{v_1}=\frac{d}{40}$

Time taken from Cambridge to oxford,  $t_2 = \frac{d}{v_2}=\frac{d}{60}$

$t_1+t_2 = d(\frac{1}{40}+\frac{1}{60})\\t_1+t_2 = d(\frac{100}{2400})\\t_1+t_2 = \frac{d}{24}$

Step 3: Find the average Velocity, $V_{avg}$

$V_{avg} = \frac{d_1+d_2}{t_1+t_2} \\ V_{avg} = \frac{2dX24}{d}\\V_{avg} = 48Km/hr$

Therefore the average speed is 48 Km/hr

#SPJ2