I was wondering why my previous post was deleted (I attached the same picture below).
I spent a long time drawing and labeling a diagram as well as writing out a detailed formulaic explanation.
Answer:
1.59 m/s²
Step-by-step explanation:
Acceleration = change in Velocity over time. Mathematically it is represented as Δv/Δt
Given values in the question:
- Final velocity 203.0 km/h which is equal to (203/3.6)m/s = 56.39 m/s
- distance 2000 m
- Initial velocity is 0 m/s
<em>To use the equation for acceleration time is needed.</em>
<em>velocity= distance/time</em>
(203/3.6) = 2000/time <em>mltiplying both sides by time</em>
(203/3.6) x time = 2000 <em>making time subject of the formula</em>
time = 2000/(203/3.6)
time = 35.47 seconds
Substituting into the acceleration equation
acceleration =Δv/Δt
=(56.39- 0)/35.47
= 56.39/35.47
=1.59 m/s²
Answer:
the 4th one
Step-by-step explanation:
Answer:
Average rate of change over the interval 2<= x <= 5:
y = 3x + 5: 3
y = 3x^2 + 1: 21
y = 3^x: 78
<u />
Step-by-step explanation:
2<= x <= 5
Average rate of change over the interval 2<= x <= 5:
<u>y = 3x + 5</u>
y(5) = 3(5) + 5 = 20
y(2) = 3(2) + 5 = 11
Average rate of change = (20 - 11)/(5-2) = 9/3 = <u>3</u>
<u />
<u>y = 3x^2 + 1</u>
y(5) = 3(5^2) + 1 = 75 + 1 = 76
y(2) = 3(2^2) + 1= 13
Average rate of change = (76 - 13)/(5-2) = 63/3 = <u>21</u>
<u />
<u>y = 3^x</u>
y(5) = 3^5 = 243
y(2) = 3^2 =9
Average rate of change = (243-9)/(5-2) = 234/3 =<u> 78</u>
Answer:
C
Step-by-step explanation:
