I am having problem with this question. Would need some help on the detail expla

Question

I am having problem with this question. Would need some help on the detail explaination and working on this problem sum. Thanks !

2)

A cat is sleeping on the floor in the middle of a 3.0 m wide room when a barking dog enters with a speed of 1.50 m/s. As the dog enters, the cat immediately accelerates at 0.85 m/s2 toward an open window on the opposite side of the room. The dog is a bit startled by the cat and begins to slow down at 0.10 m/s2 as soon as it enters the room. How far is the cat in front of the dog as it leaps through the window?

Explanation / Answer

Given

cat is at the middle of the room of wide 3 m that is at 1.5 m from the window

initial speed of the dog is u_d = 1.5 m/s ,

aceleration of the cat is a_c = 0.85 m/s2

accelerationof the dog is a_d = 0.1 m/s2

using the equations of motions as the accelerations of both are moving with constant acceleration so

now the time taken by the cat to reach window is s = ut + 0.5 at^2

t = sqrt(2*s/a) = sqrt(2*1.5/0.85) = 1.87867 s

dog initial velocity is u_d = 1.5 m/s

acceleration of the dog is a_d = -0.1 m/s^2

now the distance covered by the dog in time 1.87867 s is  

s = ut - 0.5*a*t^2

s = 1.5*1.87867 - 0.5*0.1*1.87867^2

s = 2.64153 m

by the time the cat leaves the window,the dog reached a distance of 2.64153 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.