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

(a). The acceleration is 35.5 m/s².

(b). The distance is 0.214 m.

Step-by-step explanation:

Given that,

Speed = 3.9 m/s

Time = 0.11 s

We need to calculate the acceleration

Using formula of acceleration

`a = (v_(f)-v_(i))/(t)`

Put the value into the formula

`a =(3.9-0)/(0.11)`

`a=35.5\ m/s^2`

We need to calculate the distance

Using equation of motion

`s = ut+(1)/(2)at^2`

Put the value in the equation

`s =0+(1)/(2)*35.5*(0.11)^2`

`s=0.214\ m`

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

(b). The distance is 0.214 m.

Answer 2

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