Answer:
6.18 m/s
Explanation:
Given that a ferry approaches shore, moving north with a speed of 6.1 m/s relative to the dock. A person on the ferry walks from one side of the ferry to the other, moving east with a speed of 1.0 m/s relative to the ferry Part. What is the speed of the person relative to the dock?
The speed of the person can be calculated by using pythagorean theorem.
Let the speed of the person = S
S^2 = 6.1^2 + 1^2
S^2 = 38.21
S = sqrt ( 38.21)
S = 6.18 m/s
Therefore, the speed of the person relative to the dock is 6.18 m/s
From the information in the question, the velocity of the particle is 24 ms-1.
Let us recall that the velocity of a body is obtained as the first derivative of distance covered with respect to time. Hence;
ds/dt = v
Now, we have to differentiate (t^3- 3t^2 + 2)m with respect to t and we have;
d(t^3- 3t^2 + 2)m/dt = 3t^2 - 6t
Now at t= 4s, the velocity of the particle is;
3(4)^2 - 6(4) = 24 m/s
The acceleration of the particle is the second derivative of distance with respect to time;
d^2s/dt^2 = d^2 3t^2 - 6t/dt^2 = 6t
Substituting t = 4s
a = 24 ms-2
Learn more: brainly.com/question/24486060
Larva, pupa, and egg are insects life cycle
Spore, seed, fruit, and cone are animal life cycle
Answer:
final speed of the bottle after t = 3 s is 29.43 m/s
Explanation:
As we know that the motion of the water bottle is free fall
As we know that the initial speed of the bottle is zero
so we have
now we know that
t = 3 s
now we have
