Two students walk in the same direction along a straight path, at a constant speed—one at 0.90 m/s and the other at 1.90 m/s. a. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away? b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

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mvelazcoa mvelazcoa · Answer 1 · 2019-10-05T06:12:01+02:00

Answer:

a)456.15s b)564.3m

Step-by-step explanation:

Let be student 1 the one with V1=0.90m/s as his speed, T1 his time and X1 his distance.

Let be then student 2 the one with V2=1.90m/s as his speed, T2 his time and X2 his distance.

For both

[tex]V = \frac{X}{T}[/tex]

because of their constant speed.

[tex]X1 = X2 = 780m [/tex]

then,

[tex]V1= \frac{X1}{T1} \\ T1 = \frac{X1}{V1} \\ T1 = \frac{780m}{0.9 m/s} = 866.67s [/tex]

and

[tex]V2= \frac{X2}{T2} \\ T2 = \frac{X2}{V2} \\ T2 = \frac{780m}{1.9 m/s} = 410.52s \\ [/tex]

then the second students will arrive 866.67-410.52=456.15 seconds sooner.

to get 5.50min =60s . 5.50=330 seconds sooner T2+330s=T1 and X1=X2

[tex]V2= \frac{X2}{T2} \\ T2 \times V2= X2 \\ [/tex]

[tex]V1= \frac{X1}{T1} \\ T1 \times V1= X1 \\ [/tex]

[tex]V1 \times T1=V2 \times T2 \\ V1 \times (T2 + 330s)=V2 \times T2 \\ 0.9 m/s \times (T2+330s) = 1.9 m/s \times T2 \\ [/tex]

[tex]0.9m/s \times 330s=(1.9m/s-0.9m/s) \times T2 \\ 297m = 1m/s \times T2 \\ 297s = T2 \\ [/tex]

[tex]T2 \times V2 = X2 \\ 297s \times 1.9m/s= 564.3m \\ [/tex]

they have to make 564.3m so the faster kid gets 5.50 minutes sooner