A particle was at rest from 9:00am . It moved at a uniform speed of 10km/h from 9:30am to10:00am. Find the average speed between a)9:00 am and 10:00am b) 9:15 and 10:00am

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Solution

Distance travelled between 9:30 to 10:00 i.e in 30 min or in 0.5 hr = speed×time
distance = 10×0.5 = 5 m

Avg speed between 9:00 to 10:00 i.e. in 1 hr
= total distance travelled / total time taken
= 5/1
= 5 km/hr

Time between 9:15 to 10:00 = 45 min = 3/4 hr
$a v g s p e e d = \frac{5}{\frac{3}{4}} = \frac{20}{3} = 6.66 k m / h r$

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A particle was at rest from 9:00am . It moved at a uniform speed of 10km/h from 9:30am to10:00am. Find the average speed between a)9:00 am and 10:00am b) 9:15 and 10:00am

Distance travelled between 9:30 to 10:00 i.e in 30 min or in 0.5 hr = speed×time distance = 10×0.5 = 5 m Avg speed between 9:00 to 10:00 i.e. in 1 hr = total distance travelled / total time taken = 5/1 = 5 km/hr Time between 9:15 to 10:00 = 45 min = 3/4 hr avg speed = 534 = 203 = 6.66 km/hr