Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. when you increase your speed by 8.2 mi/h , the time to go one mile decreases by 12 s . what is your original speed?

Your original speed was 45.668 mph.

To find seconds per mile, divide 3600 by speed in miles per hour. Let x represent the original speed. Then 3600/x is the original time per mile.

The new time per mile is 3600/(x +8.2), and that is 12 seconds less than the original time per mile:

... 3600/(x +8.2) = 3600/x -12

Multiplying by the product of the denominators, we have ...

... 3600(x) = 3600(x +8.2) -12(x)(x +8.2)

Dividing by 12 and putting this into standard form, we have

... x^2 +8.2x -300(8.2) = 0

... x = (-8.2 +√(8.2² -4(-300·8.2)))/2 = -4.1 +√2476.81 . . . . solved using the quadratic formula

... x ≈ 45.6676 . . . mi/h

Your original speed was about 45.67 mph.

Check

Your original time per mile was about 3600/45.66756 ≈ 78.8306 seconds. Your shorter time per mile is about 3600/53.86756 ≈ 66.8306 seconds, which is 12 seconds less.

no. dot product   =0 when vecs at right angles

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s