Question

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

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