a key is dropped from a height of 30 m , If its distance from the ground at t seconds after the release is given by the position function d (t) = 30-5t², find
a.its height velocity and acceleration after 1 second.
b. the time it reached the ground.
c. its position after 1 second
If d(t) = 30 - 5t^2 m, then the velocity is
v(t) = -10t m/sec , and the acceleration is
a(t) = -10 m/sec^2
(each is the derivative of the previous function)
a) when t = 1
distance = 30 - 5(1^2) = 25 m
velocity = -10(1) = -10 m/s
acceleration = -10 m/s^2
b) when it reaches the ground, d(t) = 0, so
0 = 30 - 5t^2
t^2 = 6
t = √6 or appr. 2.4 seconds
c) 25 m above the ground , already answered in a)
a. To find the height velocity, we need to take the derivative of the position function with respect to time:
d'(t) = -10t
Plugging in t = 1, we get:
d'(1) = -10(1) = -10 m/s
To find the acceleration, we need to take the derivative of the velocity function:
d''(t) = -10
At t = 1, the acceleration is:
d''(1) = -10 m/s²
b. To find the time it takes for the key to reach the ground, we set the position function equal to 0 and solve for t:
30 - 5t² = 0
Rearranging the equation, we get:
5t² = 30
Divide both sides by 5:
t² = 6
Taking the square root of both sides, we get:
t = √6 or t = -√6
Since time cannot be negative in this case, the key reaches the ground at t = √6 seconds.
c. To find the position after 1 second, we simply plug t = 1 into the position function:
d(1) = 30 - 5(1)² = 30 - 5 = 25 meters
Therefore, the key's position after 1 second is 25 meters above the ground.
To find the height velocity and acceleration after 1 second, we need to take the derivative of the position function d(t) = 30 - 5t² with respect to time.
a. Height Velocity:
To find the height velocity, we take the derivative of the position function:
v(t) = d'(t) = -10t
To find the height velocity after 1 second, we substitute t = 1 into the velocity function:
v(1) = -10(1) = -10 m/s
Therefore, the height velocity after 1 second is -10 m/s.
b. Time it reached the ground:
To find the time when the key reaches the ground, we set d(t) = 0 and solve for t:
30 - 5t² = 0
Solving this quadratic equation, we get two possible solutions: t = 0 and t = √6.
However, t = 0 represents the initial position where the key was dropped from, so the valid solution is t = √6.
Therefore, the key reaches the ground at approximately t = √6 seconds.
c. Position after 1 second:
To find the position after 1 second, we substitute t = 1 into the position function:
d(1) = 30 - 5(1)² = 30 - 5 = 25 meters
Therefore, the position of the key after 1 second is 25 meters.
To find the height velocity and acceleration of the key after 1 second, we need to differentiate the position function with respect to time.
a) Height Velocity after 1 second (t = 1):
To find the height velocity, we differentiate the position function with respect to time:
v(t) = d'(t) = -10t
In this case, we substitute t = 1 into the velocity function:
v(1) = -10(1) = -10 m/s
Therefore, the height velocity of the key after 1 second is -10 m/s.
b) Time it reached the ground:
To find the time the key reached the ground, we need to find the time when the height is 0. We set the position function equal to zero and solve for t:
d(t) = 0
30 - 5t² = 0
Simplifying the equation:
5t² = 30
t² = 6
t = √6 or approximately 2.45 seconds
Therefore, the key reaches the ground after approximately 2.45 seconds.
c) Position after 1 second (t = 1):
To find the position after 1 second, we substitute t = 1 into the position function:
d(1) = 30 - 5(1)²
d(1) = 30 - 5(1)
d(1) = 30 - 5
d(1) = 25 m
Therefore, the position of the key after 1 second is 25 meters above the ground.