His displacement ... the distance between his start point and
his end point ... was 10 miles. But we don't know what route
he followed on the way, so we don't know how much distance
he actually walked or drove. It couldn't be less than 10 miles,
but it could be any distance more than 10 miles.
I think the answer to this problem is b.
Answer: The First one is for x is x = 4/3 + 2t/3 and for t is t = -2 + 3x/2
And the second one is for x is x = -t/2 + 5/2 and for t is t= -2x + 5
Given
balloon rises straight up at a rate of 120 ft per min.
balloon is tracked from a rangefinder on the ground at point p, which is 400 ft from the release point q of the balloon.
d = the distance from the balloon to the rangefinder and t= the time in mins
Express d as a function of t.
To proof = With the help the diagram as shown in below .
balloon rises straight up at a rate of 120 ft per min.
thus RQ = 120t
rangefinder on the ground at point p, which is 400 ft from the release point q of the balloon.
thus PQ = 400 ft
Let PR =d
by using the pythagoras theorem
we have
RQ² + PQ² = PR²
now putting the value
we have
(120t)² + 400² = d²
= d
= d
= d
ft
hence d is express in the term of t
Hence proved