Question

problem 1. a speeding train travels a displacement of 1100 meters in 12 s before coming to a complete stop. calculate the acceleration of the train.
problem 2. a speeding train travels a displacement of 1800 meters in 18 s before coming to a complete stop. calculate the acceleration of the train.
problem 3. an ambulance begins from rest and accelerates at a rate of 1.5 m/s² for 20 s. calculate the displacement of the ambulance.
problem 4. a gorilla accelerates at a rate of -1 m/s² towards a bunch of bananas. the gorilla runs for 25 s before coming to a complete stop to eat the bananas. calculate the displacement of the gorilla.
problem 5. a runner decelerates at a rate of -1.2 m/s². if they run for 48.6 m before coming to a complete stop, determine the amount of time that they were slowing down.

Explanation:

Step1: Identify the kinematic - equation

For problems involving displacement ($\Delta x$), initial velocity ($u$), time ($t$), and acceleration ($a$), when the train comes to a stop, the final velocity $v = 0$. We can use the equation $\Delta x=ut+\frac{1}{2}at^{2}$ and $v = u+at$ (where $v = 0$ gives $u=-at$). Substituting $u=-at$ into $\Delta x=ut+\frac{1}{2}at^{2}$ gives $\Delta x=-at\times t+\frac{1}{2}at^{2}=-\frac{1}{2}at^{2}$.

Step2: Solve Problem 1

We know that $\Delta x = 1100$ m and $t = 12$ s. From $\Delta x=-\frac{1}{2}at^{2}$, we can solve for $a$. Rearranging the formula for $a$ gives $a=-\frac{2\Delta x}{t^{2}}$. Substitute $\Delta x = 1100$ m and $t = 12$ s into the formula:
\[a=-\frac{2\times1100}{12^{2}}=-\frac{2200}{144}\approx - 15.28\ m/s^{2}\]

Step3: Solve Problem 2

Given $\Delta x = 1800$ m and $t = 18$ s. Using $a=-\frac{2\Delta x}{t^{2}}$, substitute the values:
\[a=-\frac{2\times1800}{18^{2}}=-\frac{3600}{324}\approx - 11.11\ m/s^{2}\]

Step4: Solve Problem 3

The ambulance starts from rest, so $u = 0$ m/s, $a = 1.5$ m/s², and $t = 20$ s. Using the equation $\Delta x=ut+\frac{1}{2}at^{2}$, since $u = 0$, we have $\Delta x=\frac{1}{2}\times1.5\times20^{2}=\frac{1.5}{2}\times400 = 300$ m

Step5: Solve Problem 4

The gorilla has $u$ (unknown), $a=-1$ m/s² and $t = 25$ s, and $v = 0$. First, from $v = u+at$, we get $u=-at= -(- 1)\times25=25$ m/s. Then using $\Delta x=ut+\frac{1}{2}at^{2}$, substitute $u = 25$ m/s, $a=-1$ m/s² and $t = 25$ s:
\[\Delta x=25\times25+\frac{1}{2}\times(-1)\times25^{2}=25^{2}(1 - \frac{1}{2})=\frac{25^{2}}{2}=\frac{625}{2}=312.5\ m\]

Step6: Solve Problem 5

We know that $\Delta x = 48.6$ m and $a=-1.2$ m/s², $v = 0$. First, from $v^{2}-u^{2}=2a\Delta x$, since $v = 0$, we have $u^{2}=-2a\Delta x=-2\times(-1.2)\times48.6 = 2\times1.2\times48.6 = 116.64$, so $u=\sqrt{116.64}=10.8$ m/s. Then, using $v = u+at$, since $v = 0$, we can solve for $t$:
\[t=\frac{-u}{a}=\frac{- 10.8}{-1.2}=9\ s\]

Answer:

Problem 1: $a\approx - 15.28\ m/s^{2}$
Problem 2: $a\approx - 11.11\ m/s^{2}$
Problem 3: $\Delta x = 300$ m
Problem 4: $\Delta x = 312.5$ m
Problem 5: $t = 9$ s