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 `11.25 am and 11.40 am`?
(e) During a part of the journey , the car was forced to slow down to ` 12 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):}|`

Answer

(a) Graphs shown in fig. , (b) Between `10.40 am` to `10.50 am`, (c ) `26.5 km//h` , (d) ` 16 km//h` , ( e) From `E` to `F`

Explanation

(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 ` 10.40 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`.