Answer:
<h2>The answer is 5 units</h2>Step-by-step explanation:
The distance between two points can be found by using the formula
where
(x1 , y1) and (x2 , y2) are the points
From the question
The points are (9,-7) and (5, -4)
The distance between them is
We have the final answer as
<h3>5 units</h3>Hope this helps you
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>
(t -24)= 28
t - 32= 28
t= 32 +28
t= 60
t=
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)