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Bogdan [553]
3 years ago
7

I need help ASAP! It's urgent.. PLISSSSS​

Mathematics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

a) 6 mins

b) 70km/h

c) t= 45

Step-by-step explanation:

a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km

Duration

= 16 -10

= 6 minutes

b) Average speed

= total distance ÷ total time

Total time

= 24min

= (24÷60) hr

= 0.4 h

Average speed

= 28 ÷0.4

= 70 km/h

c) Average speed= total distance/ total time

Average speed

= 80km/h

= (80÷60) km/min

= 1⅓ km/min

1⅓= 28 ÷(t -24)

<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>

\frac{4}{3}(t -24)= 28

\frac{4}{3}t - 32= 28

\frac{4}{3}t= 32 +28

\frac{4}{3}t= 60

t= 6 0\div  \frac{4}{3}

t= 45

*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.

Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)

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