Two students walk in the same direction along a straight path, at a constant speed. [Solved]

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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. Thats 453.2/60 = 7.55 minutes sooner

Part b)
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, its 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)
0.9[(t 5.5)(60)] =1.9{t(60)]
There: t is in seconds but Im 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

Part b)
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

Get Answer for  Which of the following is not a valid physical security measure [Solved]

For the first part, determine how long it takes each to travel the distance, and subtract the time.
I dont know the 2nd part, sorry!!
I just posted a few Physics questions if you dont mind taking a look!

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, its 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)
0.9[(t 5.5)(60)] =1.9{t(60)]
There: t is in seconds but Im 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

1- first student need 14.44 minutes
2 2nd one need 6.84 minutes
3 they have to walk 2.36 m/s

This post is last updated on hrtanswers.com at Date : 1st of September – 2022