Answer:
The data we have is:
The acceleration is 3.2 m/s^2 for 14 seconds
Initial velocity = 5.1 m/s
initial position = 0m
Then:
A(t) = 3.2m/s^2
To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.
V(t) = (3.2m/s^2)*t + 5.1 m/s
To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)
P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t
Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.
P(14s) = (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m
So the final position is 385 meters ahead the initial position.
Answer:
Apoint on negative x-axis
Answer:
m=5/2
Step-by-step explanation:
the equation of slope is (y2-y1)/(x2-x1)
here are the points given: (0,5), (2,10)
let's label the points:
x1=0
y1=5
x2=2
y2=10
now substitute into the equation (m is the slope)
m=(10-5)/(2-0)
subtract
m=5/2
the slope is 5/2
Answer:
![\sqrt[7]{b^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B7%5D%7Bb%5E%7B15%7D%7D)
Step-by-step explanation:
=> ![b^2*\sqrt[7]{b}](https://tex.z-dn.net/?f=b%5E2%2A%5Csqrt%5B7%5D%7Bb%7D)
=> 
Where bases are same, powers are to be added.
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