If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.
Answer:let initial velocity u=14m/s
Final velocity v=20m/s
Time taken t=30
Acceleration =a
V=u +at
a= (20-14)/30
a=0.2m/s^2
Explanation:
Acceleration is the change in velocity with respect to time.
Answer:
no change in speed, therefore the body cannot be accelerated. a=0
Explanation:
When a person is accelerating his speed must change, if the speed is in the same direction as the acceleration the speed increases and if the acceleration is in the opposite direction to the speed it decreases.
In this case there is no change in speed, therefore the body cannot be accelerated.
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
<span>The Sun and all the planets revolve around Earth.</span>
