A car travels at a velocity of $80 k m / h$ during the first half of its running time and at $40 k m / h$ during the other half. Find the average speed of the car.

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Solution

For first half, distance $s_{1} = \frac{v_{1} t}{2} = \frac{80 t}{2}$

For the next half, distance $s_{2} = \frac{v_{2} t}{2} = \frac{40 t}{2}$

Average speed = $\frac{\frac{80 t}{2} + \frac{40 t}{2}}{\frac{t}{2} + \frac{t}{2}} = 60 k m h^{- 1}$

Hence, average speed is $60 k m h^{- 1}$

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the average sped of the car is:

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A car travels at a velocity of 80km/h during the first half of its running time and at 40km/h during the other half. Find the average speed of the car.

For first half, distance s1=v1t2=80t2 For the next half, distance s2=v2t2=40t2 Average speed = 80t2+40t2t2+t2=60kmh−1 Hence, average speed is 60kmh−1

A car travels at a velocity of $80 k m / h$ during the first half of its running time and at $40 k m / h$ during the other half. Find the average speed of the car.

For first half, distance $s_{1} = \frac{v_{1} t}{2} = \frac{80 t}{2}$

For the next half, distance $s_{2} = \frac{v_{2} t}{2} = \frac{40 t}{2}$

Average speed = $\frac{\frac{80 t}{2} + \frac{40 t}{2}}{\frac{t}{2} + \frac{t}{2}} = 60 k m h^{- 1}$

Hence, average speed is $60 k m h^{- 1}$