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 away?
b) How far would the students have to walk so that the faster student arrives 5.50 minutes before the slower student?
As you probably know, this is a Distance-Time-Rate question.
Consequently we must use the formula D = RT
In this question we must also watch our units of measurement
For student A, his Rate R = 0.9 and his Distance D = 780
His time T is given by T = D/R [from D=RT]
T = 780 / 0.9
T = 863.7 seconds…..[meters / meters /sec. = seconds]
For Student B, his rate is R=1.9, and Distance D also = 780
T = 780 ‘ 1.9 =410.5 seconds
The difference in time is 863.7 – 410.5 = 453.2 seconds
B arrives 453.2 seconds faster than A. That’s 453.2/60 = 7.55 minutes sooner
We again use the D = RT formula
For the faster student, B, his Rate R remains 1.9
His Distance is “D”.
His time is t
Thus for B, D = 1.9t
For student A, Distance is also “D”
His rate R remains 0.9
His time is t + 5.5 minutes
Thus for A, D = 0.9(t + 5.5)
Now at this point it is usually wise to change the minutes to seconds. I shall do this, but as
you will see in a moment, it’s not necessary in this case to do this, because the factor 60
that we use to do this will cancel out in the final calculation.
Since both students travel the same distance eventually,
D of A = D of B
0.9(t+5.5) = 1.9(t)
There: t is in seconds but I’m going to cancel both sides by 60
0,9(t+5.5) = 1.9t
0.9t + 4.95 = 1.9t
4.95 = 1.9t-0.9t
4.95 = t
B should therefore walk 1.9 X t in seconds =1.9(4.95 X 60)….meters/sec X seconds = meters
= 564.3 meters
For the first part, determine how long it takes each to travel the distance, and subtract the time.
I don’t know the 2nd part, sorry!!
I just posted a few Physics questions if you don’t mind taking a look!
1- first student need 14.44 minutes
2 2nd one need 6.84 minutes
3 they have to walk 2.36 m/s