Question

" when striking, the pike, a predatory fish, can accelerate from rest to a speed of 4.0 m/s in 0.11 s."

The question is " how far does the pike move during his strike" for part b
Part A's answer is 36.36 repeating

Answer 1

To determine the distance a pike moves during its strike, we use the kinematic equation S = 0.5at^2. Given an acceleration of approximately 36.36 m/s^2 and a time of 0.11 s, the pike moves approximately 0.22 meters during the strike.

Step-by-step explanation:

To find out how far the pike moves during its strike, we can use the kinematic equation for uniformly accelerated linear motion:

S = ut + \frac{1}{2}at^2

Where:
S = displacement (the distance moved),
u = initial velocity (which is 0 m/s since the pike starts from rest),
a = acceleration,
t = time.

Since the initial velocity u is 0, the equation simplifies to:

S = \frac{1}{2}at^2

We know the pike reaches a speed (v) of 4.0 m/s in 0.11 s. We can first calculate acceleration using the equation:

a = \frac{v - u}{t}

Inserting the known values, we get:

a = \frac{4.0 \ m/s - 0 \ m/s}{0.11 \ s} \approx 36.36 \ m/s^2

Now, we can use this acceleration to find S:

S = \frac{1}{2} × 36.36 \ m/s^2 × (0.11 \ s)^2 = 0.22 m

So, during its strike, the pike moves approximately 0.22 meters.

Answer 2

In an uniform accelerated motion, the distance covered by the fish is given by:

`S= (1)/(2)at^2`
where
a is the acceleration
t is the time

The acceleration is equal to the increase in speed of the fish divided by the time taken:

`a= (v_f-v_i)/(t)= (4 m/s-0m/s)/(0.11 s)=36.4 m/s^2`
If substitute the acceleration into the first equation, we find the distance covered by the fish:

`S= (1)/(2)at^2= (1)/(2)(36.4 m/s^2)(0.11 s)^2=0.22 m`