Question

Problem 2.26 MasteringPhysics 10 of 16 Problem 2.26 When striking, the pike, a predatory fish, can accelerate from rest to a speed of 3.9 m/s in 0.11 s Part B How far does the pike move during this strike? .

Answer 1

final velocity = initial velocity + (acceleration x time) 
3.9 m/s = 0 m/s + (acceleration x 0.11 s) 
3.9 m/s / 0.11 s = acceleration 
30.45 m/s^2 = acceleration 

distance = (initial velocity x time) + 1/2(acceleration)(time^2) 
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2) 
distance = 0.18 m

Answer 2

Answer:

(a). The acceleration is  35.5 m/s².

(b). The distance is 0.214 m.

Explanation:

Given that,

Speed = 3.9 m/s

Time = 0.11 s

We need to calculate the acceleration

Using formula of acceleration

$a = \dfrac{v_{f}-v_{i}}{t}$

Put the value into the formula

$a =\dfrac{3.9-0}{0.11}$

$a=35.5\ m/s^2$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

Put the value in the equation

$s =0+\dfrac{1}{2}\times35.5\times(0.11)^2$

$s=0.214\ m$

Hence, (a). The acceleration is  35.5 m/s².

(b). The distance is 0.214 m.