Question #21560

Question

1050 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-15T11:44:36

Yes, you may let $n$ $n$ equal their original speed.

You will need the formula for speed to solve for $n$ $n$

$speed = \frac{distance}{time}$ $"speed"="distance"/"time"$

$\Rightarrow distance = speed \cdot time$ $rArr"distance"="speed"*"time"$

Let $n = initial speed$ $n="initial speed"$

Then

$final speed = n - 5$ $"final speed"=n-5$

The distance travelled to the destination is

$d_{i} = n \cdot t = 3.5 n$ $d_i=n*t=3.5n$

The distance travelled home is the same, but given by

$d_{f} = (n - 5) t = 4 (n - 5)$ $d_f=(n-5)t=4(n-5)$

Now equate the two expressions for distance and solve for n

$d_{i} = d_{f}$ $d_i=d_f$

$\Rightarrow 3.5 n = 4 (n - 5)$ $rArr3.5n=4(n-5)$

$\Rightarrow 3.5 n = 4 n - 20$ $rArr3.5n=4n-20$

$\Rightarrow 0.5 n = 20$ $rArr0.5n=20$

$\Rightarrow n = 40 \frac{m i}{h}$ $rArrn=40" "(mi)/h$