Word Problems - Motion 2

In this question, we are asked for the speed, \(v\) on the way to school when you received a ticket.

Let's set up a table to organize the information we know. Let \(d\) represent the distance travelled in \([\text{km}]\). Let \(t\) represent the time in \([\text{hr}]\). Let \(v\) represent the speed on the way to school in \([\text{km/hr}]\):

Trip	\(d\)	\(v\)	\(t\)
School	\(d_1\)	\(v\)	\(\cfrac{10}{60}\)
Home	\(d_2\)	\(v-10\)	\(\cfrac{14}{60}\)

Notice that the speed on the way home is \(10\) less than on the way to school. Also, don't forget to convert the time to hours.

Unfortunately, we do not know the distance travelled to school or on the way home. But, we know the total distance travelled is \(25.8\;[\text{km}]\) from the odometer.

We can write this as:

\(d_1 + d_2 = 25.8\)

Remember the formula \( d = vt \). We can use this for \( d_1 \) and \( d_2 \):

School	\(d_1 = \cfrac{10}{60}v \)
Home	\(d_2 = \cfrac{14}{60}(v-10) \)

We can use substitution to get one equation with one unknown. We can then simplify this equation to determine the speed you were driving when you got your ticket:

\(d_1 + d_2 = 25.8\)

\(\cfrac{10}{60}v + \cfrac{14}{60}(v-10) = 25.8\)

\(\cfrac{10}{60}v + \cfrac{14}{60}v - \cfrac{140}{60} = 25.8\)

\(v \left(\cfrac{10}{60} + \cfrac{14}{60}\right) - \cfrac{140}{60} = 25.8\)

\(v \left(\cfrac{10}{60} + \cfrac{14}{60}\right) = 25.8 + \cfrac{140}{60}\)

\(v \left(\cfrac{24}{60}\right) = 28.13\)

\(v = \cfrac{28.13}{0.4} \)

\(v \approx 70 \; [\text{km/hr}]\)

Looks like you were driving a little fast on the way to school! On average, you drove \( 70 \; [\text{km/hr}] \) when the limit was \(60 \; [\text{km/hr}]\).