A salamander of the genus Hydromantes captures prey by launching its tongue as a projectile: The skeletal part of the tongue is shot forward, unfolding the rest of the tongue, until the outer portion lands on the prey, sticking to it. Figure 2-36
a2- .
a1- .
. .
0 10 20 30 40
shows the acceleration magnitude a versus time t for the acceleration phase of the launch in a typical situation. The indicated accelerations are a2 = 404 m/s2 and a1 = 104 m/s2. What is the outward speed of the tongue at the end of the acceleration phase?
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400
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To find the outward speed of the tongue at the end of the acceleration phase, we need to calculate the final velocity (v) using the given acceleration (a) and time (t).
The acceleration-time graph provided shows two points, a2 and a1, on the y-axis corresponding to the acceleration at different times. Let's assume that a1 is the acceleration at the beginning of the acceleration phase, and a2 is the acceleration at the end of the acceleration phase.
We can use the formula:
v = u + at
Where:
v = final velocity
u = initial velocity (usually taken as 0 in such cases)
a = acceleration
t = time
Since the initial velocity is not given, we can assume it to be 0. Therefore, the equation becomes:
v = 0 + a * t
At the end of the acceleration phase, the time (t) would be the time corresponding to a2 on the x-axis of the graph. From the graph, it appears to be around 40.
Using the given acceleration a2 = 404 m/s² and the time t = 40, we can substitute these values into the equation to calculate the final velocity:
v = 0 + 404 * 40
Calculating this, we get:
v = 16,160 m/s
Therefore, the outward speed of the tongue at the end of the acceleration phase is 16,160 m/s.
To find the outward speed of the tongue at the end of the acceleration phase, we need to integrate the acceleration curve to obtain the velocity.
Given that the acceleration versus time graph is shown in Figure 2-36, we can see that the acceleration changes from a1 = 104 m/s^2 to a2 = 404 m/s^2 over a certain period of time.
To integrate the acceleration graph, we first need to find the time interval over which the acceleration changes. From the graph, we can see that the acceleration changes from a1 to a2, between 0 and 20 milliseconds (ms).
Using the formula for acceleration, a = dv/dt, where v is the velocity and t is the time, we can integrate the acceleration graph to get the velocity graph.
∫ a dt = ∫ dv
Integrating both sides with respect to time:
∫ a dt = ∫ dv
∫ a dt = v + C
Where C is the constant of integration.
Now, let's calculate the velocity at the end of the acceleration phase:
v2 - v1 = ∫ a dt
Using the given values, a1 = 104 m/s^2, a2 = 404 m/s^2, and t = 20 ms = 0.02 s, we can substitute these values into the equation:
v2 - v1 = ∫ a dt
v2 - v1 = ∫ a dt = ∫ (a1 + [(a2 - a1)/t]t) dt
v2 - v1 = ∫ (a1 + [(a2 - a1)/t]t) dt = a1t + (a2 - a1)t/2
v2 - v1 = a1t + (a2 - a1)t/2
Now we can substitute the values for a1, a2, and t:
v2 - v1 = (104 m/s^2)(0.02 s) + (404 m/s^2 - 104 m/s^2)(0.02 s)/2
v2 - v1 = (2.08 m/s) + (300 m/s^2)(0.02 s)/2
v2 - v1 = 2.08 m/s + 6 m/s
v2 - v1 = 8.08 m/s
Therefore, the outward speed of the tongue at the end of the acceleration phase is 8.08 m/s.