This distance-time graph represents a journey made by Jo. a)What did Jo do at minute 10?
b) At what speed was Jo moving for the last 30 mins of the journey?

The slope of a straight portion of the given distance time graph gives Jo's

speed in the region where the slope is calculated.

Reasons:

(a) From the distance time graph which gives the distance from the starting point, we have;

At minute 5, Jo's distance from the starting point is 0.5 km.

At minute 10, Jo's distance from the starting point is 1 km.

At minute 15, Jo's distance from the starting point is 0.5 km.

Therefore;

Jo turned round and started a journey back to his starting point at minute 10

(b) Jo's speed in the last 30 minutes of the journey is given by the ratio of

the rise and run of the graph as follows;

60 minutes = 1 hour

30 minutes = 0.5 hour

Jo's speed the last 30 minutes of the journey = 8 km/hr.

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Jo started decelerating at the 10th minute

His average speed for the last 30 minutes of the journey is 4/30 km/minute

Step-by-step explanation:

To get what Jo did at 10 minute, we look at the graph.

At the 10th minute, we can see that we have a sharp turn from the top to the zero position at the 20th minute

so we can say that Jo started deceleration at the 10th minute

For the last 30 minutes of the journey, he travelled a distance of 4km

So, mathematically, his average speed will be;

Distance/ time = 4km/30 minutes = 4/30 km/minute

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation:

a is the right answer