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Matter In Our Surroundings

Question
CBSEENSC9005775

The following is the distance-time table of a moving car.

(i) Use a graph-paper and plot the distance travelled by the car versus time.

(ii) When was the car travelling with the greatest speed ?

(iii) What is the average speed of the car?

(iv) What is the speed between 11.25 a.m. and 11.40 a.m.?

(v) During a part of the journey, the car was forced to slow down to 12 km/h. At what distance did this happen?

Solution

(i) The distance-time graph for the motion of the car is as shown in the figure below.

(ii) The car is travelling with maximum speed between 10.40 a.m. and 10.50 a.m.
During this time the distance-time graph has maximum slope.
Here

Speed of the car =  fraction numerator 22 minus 12 over denominator 10 end fraction space equals space fraction numerator 10 over denominator bevelled 10 over 60 end fraction space equals space 60 space k m divided by h r
(iii) Average speed of car between 10.05 a.m. and 11.40 a.m.

equals space fraction numerator Total space distance over denominator Tota space time end fraction space equals space fraction numerator left parenthesis 42 minus 0 right parenthesis space km over denominator 1 space straight h space 35 space min end fraction space equals space fraction numerator 42 space km over denominator begin display style 95 over 60 end style straight h end fraction
equals space fraction numerator 42 space cross times space 60 over denominator 95 end fraction space equals space 26.52 space km divided by straight h. space

(iv) Speed between 11.25 a.m. and 11.40 a.m.
       equals space fraction numerator 4 space km over denominator 15 space min end fraction space equals space fraction numerator 4 space km over denominator begin display style 15 over 60 end style straight h end fraction space equals space 16 space km divided by straight h.

(v) Between 11.00 a.m. to 11.10 a.m., the car was forced to slow down to 12 km/h.
So, this happened at a distance of 26 km.