a cat walking at 0.25 m/s sees a mouse and accelerates uniformly at 0.40 m/s (squared) for 3.0 s. what is the cat's displacement during this time?
x=(vi)(t)+1/2a(t)^2
x=(0.25)(3)+1/2(0.40)(3)^2
x=2.55m
2.55m
2.0m
Well, let's see. A cat walking is already funny enough, but a cat accelerating towards a mouse? That's a whole new level of comedy! Alright, let's calculate this. We have the initial velocity of the cat, which is 0.25 m/s, and the acceleration, which is 0.40 m/s². We also have the time, which is 3.0 seconds. To find the displacement, we can use the equation:
displacement = initial velocity * time + 0.5 * acceleration * (time squared)
So, plugging in the values we have:
displacement = 0.25 m/s * 3.0 s + 0.5 * 0.40 m/s² * (3.0 s)²
Now, let's do the math. *Calculating noises* And the answer is... *drumroll* 0.975 meters! So, in those 3.0 seconds, the cat has traveled a total displacement of 0.975 meters in its quest to catch that elusive mouse. Good luck, Mr. Whiskers!
To find the cat's displacement during this time, we need to use the equation of motion for uniformly accelerated motion:
\[s = ut + \frac{1}{2}at^2\]
Where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(a\) is the acceleration, and
- \(t\) is the time.
Given:
- \(u\) (initial velocity) = 0.25 m/s,
- \(a\) (acceleration) = 0.40 m/s\(^2\), and
- \(t\) (time) = 3.0 s.
Now, substitute these values into the equation and calculate the displacement:
\[s = (0.25 \times 3.0) + \frac{1}{2} \times 0.40 \times (3.0)^2\]
First, calculate the initial term: \(0.25 \times 3.0 = 0.75\).
Next, calculate the second term: \(\frac{1}{2} \times 0.40 \times (3.0)^2 = 0.60 \times 9.0 = 5.40\).
Finally, add both terms together: \(0.75 + 5.40 = 6.15\).
Therefore, the cat's displacement during this time is \(6.15\) meters.