InterviewSolution
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InterviewSolution
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The following is the distance - time table of a moving car. (a) Use a graph paper to plot the distance travelled by the car versus the time . (b) What was the car travelling at the greatest speed? ( c ) What is the average speed of the car ? (d) What is the speed between 11.25amand 11.40 am? (e) During a part of the journey , the car was forced to slow down to12 km//h. At what distance did his happen? |{:("Time",,"Distance"),(10.05 am,,0 km),(10.25 am,,5 km),(10.40 am,,12 km),(10.50 am,,22 km),(11.00 am,,26 km),(11.10 am,,28 km),(11.25am,,38 km),(11.40 am,,42 km):}| |
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Answer» (b) As the slope of distance - time graph represents speed , therefore , the line with maximum slope will represents maximum speed . As is clear from Fig , line `CD` represents maximum speed between ` 10.40 am` to `10.50 am`. ( c ) Total distance travelled from ` 10.05` to `11.40 am , s = 42 km` Total time taken , `t = 11.40 - 10.05` ` = 1 h 35 m` ` =( 95)/( 60) h` Average speed , `v_(av) = ("total distance")/("time taken") =( 42 km)/( 95// 60 h) = 26.5 km//h` (d) Between `11.25 am` and `11.40 am` , distance travelled `= 42 - 38 = 4 km` `"speed" = ("distance")/("time")= ( 4 km)/( 0.25 h) = 16 km//h` (e) Slowing down must give the car minimum speed `( = 12 km//h)` . This is represented by line `EF`. For this part, distance covered `= 11.10 - 11.00 = 10 min = (10)/(60) h = (1)/( 6) h` speed `= ("distance" )/("time") = ( 2 km)/( 1//6 h) = 12 km// h`. Thus the car was forced to slow down to `12 km //h` at distance ranging from ` 26 km` to `28 km`. |
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