Question

acceleration practice problems - with conversions!

1. a car is accelerating at a rate of 3 m/s². if it has an initial velocity of 40 m/s, what is its velocity after 0.5 minutes?
2. as a ball rolls downhill it increases in speed from 6 km/hr to 3 m/s in 5 seconds. what is the acceleration of the ball in m/s²?
3. passengers get on a train at the train station and the train begins moving from a standstill. the train eventually reaches a top speed of 30 km/hr as it underwent a constant acceleration of 1.3 m/s². how many seconds did it take the train to reach its top speed?

Explanation:

1.

Step1: Convert time to seconds

We know 1 minute = 60 seconds, so 0.5 minutes = 0.5×60 = 30 s.

Step2: Use the velocity - time formula

The formula for velocity is $v = v_0+at$, where $v_0$ is the initial velocity, $a$ is the acceleration and $t$ is the time. Given $v_0 = 40$ m/s, $a = 3$ m/s² and $t = 30$ s. Then $v=40 + 3\times30=40 + 90=130$ m/s.

Step1: Convert the initial velocity to m/s

We know that 1 km = 1000 m and 1 hour = 3600 s. So, $6$ km/h=$6\times\frac{1000}{3600}=\frac{6000}{3600}=\frac{5}{3}\approx1.67$ m/s.

Step2: Use the acceleration formula

The formula for acceleration is $a=\frac{v - v_0}{t}$, where $v$ is the final velocity, $v_0$ is the initial velocity and $t$ is the time. Given $v = 3$ m/s, $v_0=\frac{5}{3}$ m/s and $t = 5$ s. Then $a=\frac{3-\frac{5}{3}}{5}=\frac{\frac{9 - 5}{3}}{5}=\frac{\frac{4}{3}}{5}=\frac{4}{15}\approx0.27$ m/s².

Step1: Convert the top - speed to m/s

We know that 1 km = 1000 m and 1 hour = 3600 s. So, $30$ km/h=$30\times\frac{1000}{3600}=\frac{30000}{3600}=\frac{25}{3}\approx8.33$ m/s.

Step2: Use the velocity - time formula to find time

The formula $v = v_0+at$, since $v_0 = 0$ (starts from stand - still), $v=\frac{25}{3}$ m/s and $a = 1.3$ m/s². We can solve for $t$, $t=\frac{v - v_0}{a}=\frac{\frac{25}{3}-0}{1.3}=\frac{25}{3\times1.3}=\frac{25}{3.9}\approx6.41$ s.

Answer:

130 m/s

2.