I can't figure this out at all...Can someone please explain it to me?
It has been claimed that an insect is the best jumper in the animal kingdom. This insect can accelerate at 3950 m/s2 over a distance of 2.3 mm as it straightens its legs.
(a) Assuming a uniform acceleration, what is the velocity of the insect after it has accelerated through this short distance, and how long did it take to reach that velocity?
(b) How high would the insect jump if air resistance could be ignored? Note that the actual height obtained is about 0.7 m, so air resistance is important here.
Thank you so much!
Sure, I can help you with that.
(a) To find the velocity of the insect after accelerating through a distance of 2.3 mm, we can use the equation:
v^2 = u^2 + 2as
where:
v is the final velocity,
u is the initial velocity (which we assume to be zero in this case),
a is the acceleration, and
s is the distance traveled.
Given that the acceleration of the insect is 3950 m/s^2 and the distance traveled is 2.3 mm (or 0.0023 m), we can plug in these values into the equation:
v^2 = 0 + 2 * 3950 * 0.0023
Simplifying this equation, we get:
v^2 = 18.07
Taking the square root of both sides, we find:
v = 4.25 m/s (rounded to two decimal places)
So, the velocity of the insect after accelerating through this distance is 4.25 m/s.
To calculate the time it took to reach this velocity, we can use the equation:
v = u + at
Since the initial velocity is zero, this equation simplifies to:
v = at
Rearranging the equation to solve for t, we have:
t = v / a
Plugging in the values for v and a, we get:
t = 4.25 / 3950
t ≈ 0.00108 seconds (rounded to five decimal places)
Therefore, it took approximately 0.00108 seconds for the insect to reach a velocity of 4.25 m/s.
(b) To calculate the height the insect would jump if air resistance could be ignored, we can use the equation for the maximum height of a projectile:
h = (v^2) / (2g)
where:
h is the height,
v is the initial vertical velocity, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, since the insect is jumping vertically, the initial vertical velocity is equal to the final velocity attained during the acceleration phase.
Using the value of the final velocity we found in part (a), we have:
h = (4.25^2) / (2 * 9.8)
h ≈ 0.95 meters (rounded to two decimal places)
Therefore, if air resistance is ignored, the insect would jump approximately 0.95 meters high.
Keep in mind that in reality, the insect jumps only about 0.7 meters, which suggests that air resistance plays a significant role in limiting its actual jumping height.