If a car traveling at 60 mph is chasing a car traveling 50 mph and is 1/4 mile behind, how many minutes will it take the first car to catch the second?
So initially when approaching this problem I subtracted 50mph from 60mph so I can know how many miles per hour faster the car is going which I got to be 10mph. From there I know I want to do something with the .25 miles behind but I'm not quite sure. Can someone help me with how to approach this?
Let $t$ be the time from the start at which the chaser catches the leader. In $t$ time, the chaser would have traveled $60t$ and the leader would have traveled $50t$. But the initial lead of $0.25$ miles is also to be taken care of by the chaser so $$60t=0.25+50t.$$ Now solve for $t$.
The idea is that relative to car $A(chaser)$ the car $B(chasee)$ has a velocity of $-10\ mph$, and initially a distance of 0.25 $miles$ away. So relative to $A$, how long will $B$ take to cover the same?
The position of a car (from a certain starting point, let's say it's $0$ to keep things simple) can be given by the formula
x=v*t
where $v$ is its constant(!) speed in mph and $t$ is the time in h traveled since the position $0$. If the speed wasn't constant, this formula couldn't be applied.
Let's call the position of the 60 mph car $x_1$ and the position of the 50 mph car $x_2$. This gives us the following set of formulae:
$x_1=60t$
$x_2=50t$
The difference between these positions is $\frac{1}{4}=0.25$, so $x_2-x_1=0.25$. Notice the order in which I am subtracting, because the 50mph car ($x_2$) has a higher position value than the 60mph one ($x_1$), since the 50mph car has the lead on the 60mph one. Substituting the two equations above into $x_2-x_1=0.25$ we get
$50t-60t=0.25$
so after applying the basic algebraic rule $(a+b)*c=a*c+b*c$ this equals to
$(50-60)t=-10t=0.25$
and dividing both sides in the equation $-10t=0.25$ by $-10$ gives
$t=\frac{0.25}{-10}=-0.025$.
So this is the amount of hours the car will have to travel to get to the other car. The negative sign means that the car is behind the other one and can be disregarded. So just convert $0.025$ hours to minutes to get your result.
