Question

object a accelerates at 10 m/s²; object b accelerates at 5 m/s². which one will go faster? which one will take more time to reach a high speed? which one will reach 40 m/s first? if they start at rest, which one will go farther (longer distance)? which one will be 100m away sooner?

Explanation:

Step1: Recall the kinematic - equation for speed

The speed - time relationship for an object with constant acceleration starting from rest is $v = at$. For object A with $a_A=10\ m/s^2$ and object B with $a_B = 5\ m/s^2$, at any given time $t$, $v_A=10t$ and $v_B = 5t$. Since $10t>5t$ for $t>0$, object A will go faster.

Step2: Recall the kinematic - equation for time to reach a certain speed

We use the formula $v = at$, so $t=\frac{v}{a}$. To reach $v = 40\ m/s$, for object A, $t_A=\frac{40}{10}=4\ s$, and for object B, $t_B=\frac{40}{5}=8\ s$. So object A will reach $40\ m/s$ first.

Step3: Recall the kinematic - equation for distance

The distance - time relationship for an object starting from rest is $x=\frac{1}{2}at^{2}$. For a fixed time $t$, $x_A=\frac{1}{2}\times10t^{2}=5t^{2}$ and $x_B=\frac{1}{2}\times5t^{2}=2.5t^{2}$. So object A will go farther (longer distance) in the same time.

Step4: Recall the kinematic - equation for time to reach a certain distance

We use the formula $x=\frac{1}{2}at^{2}$, so $t=\sqrt{\frac{2x}{a}}$. To reach $x = 100\ m$, for object A, $t_A=\sqrt{\frac{2\times100}{10}}=\sqrt{20}\approx4.47\ s$, and for object B, $t_B=\sqrt{\frac{2\times100}{5}}=\sqrt{40}\approx6.32\ s$. So object A will be $100\ m$ away sooner.

Answer:

Object A will go faster, reach 40 m/s first, go farther (longer distance), and be 100 m away sooner.