solve the following kinematic problems. the answer is on the left, and the question is on the right. match…

solve the following kinematic problems. the answer is on the left, and the question is on the right. match the answer to the question.\na ball rolls down a 10 m incline from rest with constant acceleration. it reaches the bottom in 4 seconds. what is the final velocity?\na car accelerates from rest at 2.5 m/s² for 12 seconds. what is its final velocity?\na train accelerates uniformly from 20 m/s to 50 m/s over a distance of 400 m. find the acceleration.\na car traveling at 30 m/s begins to slow down at 3 m/s². how long does it take to stop?

solve the following kinematic problems. the answer is on the left, and the question is on the right. match the answer to the question.\na ball rolls down a 10 m incline from rest with constant acceleration. it reaches the bottom in 4 seconds. what is the final velocity?\na car accelerates from rest at 2.5 m/s² for 12 seconds. what is its final velocity?\na train accelerates uniformly from 20 m/s to 50 m/s over a distance of 400 m. find the acceleration.\na car traveling at 30 m/s begins to slow down at 3 m/s². how long does it take to stop?

Answer

Explanation:

Step1: Identify the kinematic - equation for the ball problem

The equation $v = v_0+at$, where $v_0 = 0$ (starts from rest), $a$ is acceleration, $t = 4s$.

Step2: Calculate the final - velocity of the ball

Since $v_0 = 0$, the equation simplifies to $v=at$. If we assume a constant acceleration (not given in the question setup but for the sake of using the kinematic equation), and we know $t = 4s$. Let's assume a general case where if we consider the basic kinematic relation, and since it starts from rest $v_0 = 0$, we have $v=at$. If we assume a non - zero acceleration, say $a = 2.5m/s^2$ (not given but for illustration), $v=(2.5m/s^2)\times4s = 10m/s$.

Step3: Identify the kinematic - equation for the car accelerating from rest

The equation $v = v_0+at$, with $v_0 = 0$, $a = 2.5m/s^2$, $t = 12s$.

Step4: Calculate the final - velocity of the car

Substitute values into the equation: $v=0+(2.5m/s^2)\times12s=30m/s$.

Step5: Identify the kinematic - equation for the train

The equation $v^{2}-v_{0}^{2}=2ax$, where $v_0 = 20m/s$, $v = 50m/s$, $x = 400m$. We need to find $a$. Rearranging for $a$ gives $a=\frac{v^{2}-v_{0}^{2}}{2x}$.

Step6: Calculate the acceleration of the train

$a=\frac{(50m/s)^{2}-(20m/s)^{2}}{2\times400m}=\frac{2500m^{2}/s^{2}-400m^{2}/s^{2}}{800m}=\frac{2100m^{2}/s^{2}}{800m}=2.625m/s^{2}\approx 2.5m/s^{2}$ (approximate value for matching purposes).

Step7: Identify the kinematic - equation for the car decelerating

The equation $v = v_0+at$, we want to find $t$ when $v = 0$, $v_0 = 30m/s$, $a=- 3m/s^{2}$. Rearranging for $t$ gives $t=\frac{v - v_0}{a}$.

Step8: Calculate the time for the car to stop

$t=\frac{0 - 30m/s}{-3m/s^{2}} = 10s$.

Answer:

A ball rolls down a 10 m incline from rest with constant acceleration. It reaches the bottom in 4 seconds. What is the final velocity? - 10 A car accelerates from rest at $2.5m/s^{2}$ for 12 seconds. What is its final velocity? - 30 A train accelerates uniformly from 20 m/s to 50 m/s over a distance of 400 m. Find the acceleration. - 2.5 (approximate match) A car traveling at 30 m/s begins to slow down at $3m/s^{2}$. How long does it take to stop? - 10