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The following is the distance time table of a moving car : time. distance10.05a.m. 0 km10.25a.m 5 km10.40a.m. 12km10.50a.m. 22km11.00a.m. 26km11.10a.m. 28km11.25a.m. 38km11.40a.m. 42km(1) use graph paper and plot the distance traveled by the car versus time(2) when was the car travelling at the greatest speed(3) what is the average speed of the car(4) what is the speed between 11.25a.m. and 11.40a.m.(5) during a part of the journey the car was forced to slow down to 12 km at what distance did this happen |
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Answer» as followStep-by-step explanation:(a) GRAPHS shown in fig. , (b) Between 10.40am to 10.50am, (c ) 26.5km/h , (d) 16km/h , ( e) From E to F Solution (a) Figure represents the distance - time graph of the car . Time is taken along X - axis and distance is taken along Y - axis. (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 10.40am to 10.50am. ( c ) Total distance travelled from 10.05 to 11.40am,s=42km Total time taken , t=11.40−10.05 =1h35m =9560h Average speed , vav=total distancetime taken=42km95/60h=26.5km/h (d) Between 11.25am and 11.40am , distance travelled =42−38=4km speed=distancetime=4km0.25h=16km/h (e) Slowing down MUST give the car MINIMUM speed (=12km/h) . This is REPRESENTED by line EF. For this part, distance covered =11.10−11.00=10min=1060h=16h speed =distancetime=2km1/6h=12km/h. Thus the car was forced to slow down to 12km/h at distance ranging from 26km to 28km. |
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