A car moves with a speed of $40 k m / h$ for $15$ minutes and then with a speed of $60 k m / h$ for the next $15$ minutes. The total distance covered by the car is:

100 k m

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25 k m

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15 k m

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10 k m

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Solution

The correct option is B $50 km$
$< v >= \frac{(D_{1} + D_{2})}{t_{1} + t_{2}} = \frac{(40 (\frac{15}{60}) + 60 (\frac{15}{60}))}{((\frac{15}{60}) + (\frac{15}{60}))}$
$= \frac{100}{2} = 50 k m$

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A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

A car moves with a speed of 40 km/hr for 15 minutes and then with a speed of 60 km/hr for the next 15 minutes. What is the total distance covered by the car? Also, find the average speed of the car. [5 MARKS]

Q. A car moves with a speed of 60km/hr for 15 minutes and then with a speed of 80km/hr for next 15 minutes. Find the total distance covered by the car.

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A car moves with a speed of 40km/h for 15 minutes and then with a speed of 60km/h for the next 15 minutes. The total distance covered by the car is:

The correct option is B 50km50km50km<v>=(D1+D2)t1+t2=(40(1560)+60(1560))((1560)+(1560))=1002=50km

A car moves with a speed of $40 k m / h$ for $15$ minutes and then with a speed of $60 k m / h$ for the next $15$ minutes. The total distance covered by the car is:

The correct option is B $50 km$
$< v >= \frac{(D_{1} + D_{2})}{t_{1} + t_{2}} = \frac{(40 (\frac{15}{60}) + 60 (\frac{15}{60}))}{((\frac{15}{60}) + (\frac{15}{60}))}$
$= \frac{100}{2} = 50 k m$