Question

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles, find the total distance traveled in the two days.

Answer 1

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$, distance of $d_1$, v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$.

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$, which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$, and the total distance is:

$T = d_1 + d_2 = d_1 + d_1 - 20 = 2d_1 - 20$

From the first equation:

$d_1 = 40t_1$

$t_1 = \frac{d_1}{40}$

Replacing in the second equation:

$60 = \frac{d_1 - 20}{t_1 - 2}$

$d_1 - 20 = 60t_1 - 120$

$d_1 - 20 = 60\frac{d_1}{40} - 120$

$d_1 = \frac{3d_1}{2} - 100$

$d_1 - \frac{3d_1}{2} = -100$

$-\frac{d_1}{2} = -100$

$\frac{d_1}{2} = 100$

$d_1 = 200$

Thus, the total distance is:

$T = 2d_1 - 20 = 2(200) - 20 = 400 - 20 = 380$

The total distance traveled in two days was of 380 miles.

For the relation between velocity, distance and time, you can take a look here: brainly.com/question/14307500