A mouse is running away from a cat at. 75 m/s and, at 1.25 m from his mouse hole, he accelerates at a constant rate of. 25 m/s^2 until he reaches his hole safely. How long a time was required for him to cover te last 1.25m?

1.4

To find the time required for the mouse to cover the last 1.25 m, we can use the kinematic equation:

\[ s = ut + \frac{1}{2}at^2 \]

Where:
- s is the displacement (1.25 m),
- u is the initial velocity (75 m/s),
- a is the acceleration (0.25 m/s^2),
- t is the time we want to find.

Rearranging the equation, we have:

\[ t^2 + (2u/a)t - 2s/a = 0 \]

Plugging in the given values:

\[ t^2 + (2 * 75 / 0.25)t - 2 * 1.25 / 0.25 = 0 \]

Simplifying the equation, we get:

\[ 4t^2 + 600t - 20 = 0 \]

We can solve this quadratic equation using the quadratic formula:

\[ t = \frac{-b ± √(b^2 - 4ac)}{2a} \]

In this case, a = 4, b = 600, and c = -20.

\[ t = \frac{-600 ± √(600^2 - 4 * 4 * -20)}{2 * 4} \]

Calculating the values within the square root:

\[ t = \frac{-600 ± √(360000 - (-320))}{8} \]

\[ t = \frac{-600 ± √(360000 + 320)}{8} \]

\[ t = \frac{-600 ± √(360320)}{8} \]

\[ t = \frac{-600 ± 600.267}{8} \]

Taking the positive value:

\[ t = \frac{-600 + 600.267}{8} \]

\[ t = \frac{0.267}{8} \]

\[ t \approx 0.0334 \, \text{s} \]

Therefore, it took approximately 0.0334 seconds for the mouse to cover the last 1.25 m.

To find the time required for the mouse to cover the last 1.25 meters, we can use the kinematic equation:

\[d = v_i t + \frac{1}{2} a t^2\]
Where:
- \(d\) is the displacement (1.25 m in this case)
- \(v_i\) is the initial velocity (75 m/s in this case)
- \(a\) is the acceleration (0.25 m/s\(^2\) in this case)
- \(t\) is the time we want to find

Since we want to know the time it takes for the mouse to cover the last 1.25 meters, we can rearrange the equation and solve for \(t\):
\[t = \frac{-v_i \pm \sqrt{v_i^2 - 4 \cdot \frac{1}{2} \cdot a \cdot -d}}{2 \cdot \frac{1}{2} \cdot a}\]

Plugging in the given values, we have:
\[t = \frac{-75 \pm \sqrt{75^2 - 4 \cdot \frac{1}{2} \cdot 0.25 \cdot -1.25}}{2 \cdot \frac{1}{2} \cdot 0.25}\]

Simplifying further:
\[t = \frac{-75 \pm \sqrt{5625 + 0.625}}{0.25}\]
\[t = \frac{-75 \pm \sqrt{5625.625}}{0.25}\]

Using the quadratic formula, we can find the square root of the discriminant:
\[t = \frac{-75 \pm 75.03}{0.25}\]

Splitting it into two equations:
\[t_1 = \frac{-75 + 75.03}{0.25}\]
\[t_2 = \frac{-75 - 75.03}{0.25}\]

Simplifying further:
\[t_1 = \frac{0.03}{0.25}\]
\[t_2 = \frac{-150.03}{0.25}\]

Calculating:
\[t_1 \approx 0.12 \, \text{s}\]
\[t_2 \approx -600 \, \text{s}\]

Since time cannot be negative, we can conclude that the mouse took approximately 0.12 seconds to cover the last 1.25 meters.

use the formula:

S = ut +(1/2)*a*t^2

S - distance covered =1.25m
u - initial velocity = 75m/s^2
a - acceleration = 25m/s^2
t - time taken to cover the distance