Physics 1: Problem of the Day Archive

Problem Daily Day 2
September 13, 2020

The Problem

Jennifer jogs at constant speed of of 3.2 meters per second to her neighbor's house down the block, which is 530 meters away. How long did it take her to get there?

The Answer

The secret to solving word problems is to G.U.E.S.S.!

  • G stands for Given. Write down all the information given that will help you solve the problem. This can include a picture, unit conversions if needed, or anything else in the problem that can help you solve it.
  • U stands for Unknown. Write down what the problem asks you to find. I like to write it like "$x = ?$"
  • E stands for Equation. Look at the Givens and the Unknown, and try to think of a formula or equation that you have enough information to fill out everything except the unknown.

Before we continue, take a look and notice that you have translated to word problem to a regular problem. Yay!

  • S and S stand for Substitute and Solve, but depending on the problem and your math skills, they don't have to be in that order.

Once you are done GUESSing, you will have the answer to your word problem. Now let's see how it works!

Given

Reading the probelem gives us this information:

  • $v_{i} = 3.2 \frac{\text{m}}{\text{s}}$ because "Jennifer jogs at a constant speed of 3.2 meters per second"
  • $v_f = 3.2 \frac{\text{m}}{\text{s}}$ because "Jennifer jogs at a constant speed of 3.2 meters per second"
  • $s = 530 \text{m}$ because "530 meters away"
  • $a = 0 \frac{\text{m}}{\text{s}^2}$ because "jogs at a constant speed"

The only quantity that isn't explicitly stated, and has to be inferred, is the acceleration. We look for the context clue of "constant speed", which we should interpret as "zero acceleration".

To see why this is, we can look at the definition for acceleration: $a \equiv \frac{v_f - v_i}{t_f - t_i}$. Since velocity (or speed in this case) is constant, then $v_f = v_i$, which implies that $v_f - v_i = 0$.

So $a = \frac{0 \frac{\text{m}}{\text{s}}}{t_f-t_i} = 0$

Unknown

The question at the end of the problem usually tells us what the unknown is.

  • $t = ?$ because "How long did it take her to get there?"

Equation

This problem is in the topic of 1-D Motion, or Kinematic Equations. There are only 3 equations relating to this topic; one equation relates our givens and unknown.

  • $v_f = v_i + a t$
  • $s = v_i t + \frac 1 2 a t^2$ ← ← ← ← ← This one!
  • $2 a s = v_f^2 - v_i^2$

The first two equations contains the unknown, $t$.

So we can ignore the third equation for this problem.

Now, how do we choose between the remaining two equations? Well, the second equation contains the other given variables, the displacement ($s$), and the initial speed, $v_i$.

SS...

Believe it or not, at this point the problem is pretty much solved. A little bit of Algebra 1 is all that remains. There are a couple of different ways this can be done...

Method 1: Substitute, then Solve

If you are uncomfortable manipulating an equation without all the numbers, you should probably substitute first. Rewrite the equation and "plug in" the values that were given. Then solve, probably using a calculator quite a bit. (Note that the units are all SI units, so they should work out.)

$$s = v_i t + \frac 1 2 a t^2$$


$$530 = 3.2 \times t + \frac 1 2 0 t^2$$ $$530 / 3.2 = t$$
$$t = 530 / 3.2 \text{s}$$

$$t = 166 \text{s}$$

Method 2: Solve, then Substitute

If you can handle it, consider solving the symbolic equation before plugging in the numbers or grabbing a calculator. Since there are only a few equations and a few variables, there can only be a relatively small number of unique problems. Solving symbolically and then plugging in the givens is a good way to start noticing that, and also reduces your calculator usage. Here, since we have an equation that is quadratic in $t$ (that is, the variable we are solving for has a $t^2$ term in the equation), we can do one of two things at this point.

(1) We can get the general solution for $t$ by using the quadratic formula

(2) We can notice that the quadratic term (that is, the $t^2$ term) will be zero because it is multiplied by a zero acceleration. For now, we will choose this second option, but we will show you the general solution in a later problem.

$$s = v_i t + \frac 1 2 a t^2$$$$s = v_i t + \frac 1 2 0 t^2$$$$s = v_i t$$$$\frac{s}{v_i} = t$$$$t = \frac{s}{v_i}$$$$t = \frac{530 \text{m}}{3.2 \frac{\text{m}}{\text{s}}}$$$$t = 530 / 3.2 \text{s}$$

$$t = 166 \text{s}$$

The End! Problem solved!