Answer:
Step-by-step explanation:
To find : Acceleration in first 15 min . Distance between two cities Average speed of journey
Solution:
Each horizontal block is 1/8 hr = 7.5 min
Each vertical block is 10 km/hr
Time Velocity km/hr
0 Min ( 0 hr) 0
15 Min (1/4 hr) 50
45 Min (3/4 hr) 50
60 MIn ( 1 hr) 100
90 Min ( 3/2 hr) 100
120 Min ( 2hr) 0
Acceleration in first 15 min (1/4 hr) = (50 - 0)/(1/4 - 0) = 50/(1/4)
= 200 km/h²
Distance between two cities
= (1/2)(0 + 50)(1/4 - 0) + 50 * (3/4 - 1/4) + (1/2)(50 + 100)(1 - 3/4) + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)
= 25/4 + 25 + 75/4 + 50 + 25
= 125
Distance between two cities = 125 km
Average Speed of journey = 125/2 = 62.5 km/hr
Acceleration in first 15 min = 200 km/h²
Distance between two cities = 125 km
Average Speed of journey = 62.5 km/hr
Hope this helps..
Answer is b
–y^–4 = – 1/y^4
hope that helps
16/20
Simplified to 8/10
Again simplified to 4/5.
4(x-5)=24 would be your answer.
