Chapter 2: Q68P (page 36)

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 2-39 shows the acceleration magnitude a versus time t for the acceleration phase of the launch in a typical situation. The indicated accelerations are $a_{2} = 400 m / s^{2} and a_{1} = 100 m / s^{2}$ . What is the outward speed of the tongue at the end of the acceleration phase?

Short Answer

Expert verified

The outward speed of the tongue at the end of the acceleration phase is 5 m/s .

Step by step solution

Step 1: Given data

The plot of $a (\frac{m}{s^{2}})$ vsrole="math" localid="1656159919636" $t (ms)$ . The acceleration $a_{1} = 100 \frac{m}{s^{2}}$ and $a_{2} = 400 \frac{m}{s^{2}} .$ .

Area under the curve

The calculation of the region bounded by the curve within the given graph is called the area under the curve. The velocity can be determined by the graphical integration of acceleration versus time.The area under the curve can be divided into three parts, from 10 ms to 20 ms, from 20 ms to 30 ms, and from 30 ms to 40 ms.

Calculating the outward speed of the tongue

The area under the acceleration curve, which is the area under the tongue of the salamander given by

$V = \frac{1}{2} \times 10^{- 2} s \times 100 \frac{m}{s^{2}} + \frac{1}{2} \times 10^{- 2} \times (100 \frac{m}{s^{2}} + 400 \frac{m}{s^{2}} + \frac{1}{2} \times 10^{- 2} \times 400 \frac{m}{s^{2}} = 5 m / s$