Question

1. 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) When 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 this
happen?
Time
Distance
10-05 am
10-25 am
10-40 am
10-50 am
11.00 am
11.10 am
11-25 am
11-40 am
0 km
5 km
12 km
22 km
26 km
28 km
38 km
42 km​

Answer 1

Answer:

Sorry for photo that is not clear

Answer 2

Using Distance-time graph

(a) Refer to the attached image for the plot.

(b)  From the graph, During the time period from 10.40am to 10.50 it has a highest Slope. Therefore, The car travels at greatest speed between 10.40am to 10.50.

(c) Average Speed = $\frac{Total \ distance \ travelled}{Total \ time \ taken}$

Total distance travelled = 42-0 = 42 Km.

Total time taken = 11.40 - 10.05= 1 hour 35 min= $1+\frac{35}{60} = \frac{95}{60} \ hours$

∴ Average Speed = $\frac{42}{\frac{95}{60} }$

$=\frac{42 \times60}{95} \\= 26.5 \ Km/h$

(d) Speed between 11.25 am and 11-40 am  = Distance /Time

$=\frac{42-38}{0.25} \\\\=\frac{4}{0.25} \\\\= 16 \ Km/hr$

(e)During  11:00 am to 11:10am the speed of the car is

$=\frac{2}{\frac{1}{6} } \\\\= 12 \ Km/hr$

Thus the car was forced to slow down to 12km/h at distance from 26km to 28km.

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Learn More

1)Show the shape of the distance-time graph for the motion in the following cases:

(i) A car moving with a constant speed.

(ii) A car parked on a side road.

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2)This distance-time graph represents a journey made by sue. Work out how much time sue spends travelling and how much time she spends stationery

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