A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
2t3 (a) find the velocity at time t (in ft/s). v(t) = (b) what is the velocity after 1 second(s)? v(1) = ft/s (c) when is the particle at rest? t = s (smaller value) t = s (larger value)
1 answer:
The distance (ft) traveled by the particle at time t (s) is
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
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(a) 18.9 m/s
The motion of the stone consists of two independent motions:
- A horizontal motion at constant speed
- A vertical motion with constant acceleration () downward
We can calculate the components of the initial velocity of the stone as it is launched from the ground:
The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.
When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore
(b) 22.2 m/s
We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by
(1)
where
is the initial vertical velocity
is the vertical velocity at height h
is the acceleration due to gravity (negative because it is downward)
At the top of the parabolic path, , so we can use the equation to find the maximum height
So, at half of the maximum height,
And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:
And so, the speed of the stone at half of the maximum height is
(c) 17.4% faster
We said that the speed at the top of the trajectory (part a) is
while the speed at half of the maximum height (part b) is
So the difference is
And so, in percentage,
So, the stone in part (b) is moving 17.4% faster than in part (a).