## Related questions with answers

Squids can move through the water using a form of jet propulsion. Suppose a squid jets forward from rest with constant acceleration for $0.170 \mathrm{~s}$, moving through a distance of $0.179 \mathrm{~m}$. The squid then turns off its jets and coasts to rest with constant acceleration. The total time for this motion (from rest to rest) is $0.400 \mathrm{~s}$, and the total distance covered is $0.421 \mathrm{~m}$. What is the time it coasts to a stop?

Solution

VerifiedFrom, (a) now we know the acceleration of the squid during propulsion, we need to find its final speed at the end of propulsion in order to find the squid is magnitude of deceleration while coasting.

To find the final speed, knowing initial speed, distance during which it accelerated and the magnitude of acceleration, we use the following equation of motion

$v^2 = v^2_{o} + 2 a \Delta x$

where,

$v$: it the final speed of squid at the of propulsion.

$v_{o}$: squid it initial speed which is $\textbf{zero}$.

$a$: squid acceleration during propulsion

$\Delta x$: the distance the squid swam while propulsion $0.179$ m.

$v^2= 2 a \Delta x$

$v = \sqrt{ 2 a \Delta x } = \sqrt{2 \times 12.4 \times 0.179} =2.11 \dfrac{\mathrm{m}}{\mathrm{s}}$

Now, that we know the speed by which the squid begun coasting and given the statement that it took 0.4 second for the squid to finish the whole propulsion and coasting motion, i.e. the squid spent 0.17 out of 0.42 second propulsing and then spent the rest of time coasting.

Though, the time squid took coasting is simply $0.4-0.17 = 0.23$ seconds and also the whole motion took place in a distance of 0.421 m thus squid decelerated while coasting in a distance of $0.42-0.179= 0.242$ m in a time interval of 0.23 seconds.

Knowing the initial speed the squid begun coasting , distance and time squid took during deceleration we can find the magnitude of acceleration using the following equation of motion.

$\Delta x = v_{o} t + \dfrac{1}{2} a t^2$

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