On average, froghopper insects have a mass of 12.3 mg and jump to a height of 428 mm. The takeoff velocity is achieved as the little critter flexes its leg over a distance of approximately 2.0 mm. Assume a vertical jump with constant acceleration. How long does the jump last (the jump itself, not the time in the air) and what is the froghopper's acceleration during that time?
To find the duration of the jump and the acceleration of the froghopper during that time, we can use the equations of motion.
First, let's convert the given distances and mass to SI units:
Mass of the froghopper = 12.3 mg = 12.3 × 10^(-6) kg
Jump height = 428 mm = 428 × 10^(-3) m
Flex distance = 2.0 mm = 2.0 × 10^(-3) m
Now, let's analyze the motion of the froghopper during the jump. Since the motion is vertical, we can consider only the vertical component of the jump.
1. Find the initial velocity (v0) of the froghopper:
Using the equation of motion, v^2 = u^2 + 2as, where v = 0 (as the froghopper starts from rest), u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since we know the distance traveled (2.0 mm) and the acceleration is constant, the initial velocity (u) can be calculated as follows:
0 = u^2 + 2a(2.0 × 10^(-3))
u^2 = -2a(2.0 × 10^(-3))
u = sqrt(-2a(2.0 × 10^(-3)))
2. Find the time (t) taken for the jump:
Using the equation of motion, v = u + at, where v is the final velocity (0 when the froghopper reaches its maximum height), u is the initial velocity, a is the acceleration, and t is the time taken.
0 = sqrt(-2a(2.0 × 10^(-3))) + a*t
0 = sqrt(-4a(2.0 × 10^(-3))) + a*t
0 = a*t - sqrt(8a × 10^(-3))
Simplified: a*t = sqrt(8a × 10^(-3))
t = sqrt(8 × 10^(-3))
3. Find the acceleration (a) during the jump:
Using the equation of motion, v = u + at, substituting the value of v as 0 when the froghopper reaches its maximum height, and t as sqrt(8 × 10^(-3)):
0 = sqrt(-2a(2.0 × 10^(-3))) + a*sqrt(8 × 10^(-3))
sqrt(-2a(2.0 × 10^(-3))) = a*sqrt(8 × 10^(-3))
sqrt(-2(2.0 × 10^(-3))) = a*sqrt(8 × 10^(-3))
Simplified: sqrt(-4) = a*sqrt(8)
-4 = 8a
a = -0.5 m/s^2
Therefore, the duration of the jump is sqrt(8 × 10^(-3)) seconds, approximately equal to 0.0894 seconds. The acceleration of the froghopper during that time is -0.5 m/s^2.
Note: The negative sign indicates that the acceleration is directed opposite to the motion, which is the deceleration phase of the jump.