Question

10. A cat walking at 0.25 m/s sees a mouse and accelerates uniformly at 0.40 m/s

for 3.0 s. What is the cat's displacement during this time?

Answer 1

7.5 m

Step-by-step explanation:

A cat walking at 0.5 m/s sees a mouse and accelerates uniformly at 0.40 m/s2 for 5.0 s. What is the cat’s displacement during this time? Show your work (Use Insert/Equation).

Given: V₁ = 0.55 m/s; a = 0.40 m/s²; Δt = 5.0s

Unknown: ΔX

Equation: ΔX = V₁ Δt + ½ a(Δt)²

displacement = (initial velocity x time interval) + ½acceleration x (time interval)²

Plug in values.

(0.5 m/s x 5.0 s) + ½ 0.40 m/s² (5.0s)²

solve (show work) (rmr pemdas rules)

7.5 m

Answer 2

The displcament is 2.55m

Step-by-step explanation:

step one:

Given that the intial speed is

u= 0.25 m/s

and acceleration is

a= 0.40 m/s^2

and time t= 3seconds

step two:

Applying the formula

s=ut+1/2at^2

s= 0.25*3+ 1/2*0.4*3^3

s= 0.75+3.6/2

s= 0.75+1.8

s=2.55 m

The displacement is 2.55m