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:
p(4 successes) ≈ 7.7%
Step-by-step explanation:
p(k successes in n trials) = C(n,k)·p^k·(1-p)^(n-k)
You have n=5, k=4, p=0.4, so the probability is ...
p(4 successes) = C(5,4)·(0.4)^4·(1 -0.4)^1 = 5·0.4^4·0.6 = 0.0768
p(4 successes) ≈ 7.7%
Y = kx
4.8 = 4k
k = 4.8/4 = 1.2
Therefore, required equation = y = 1.2x
