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Question

PHYSICS HELPERS I need your help on 3 word problems help?
PHYSICS HELPERS I need your help on 3 word problems help?
QUESTION !
Two towns, A and B, are 200 miles apart. Car 1 starts from town A and at the same time, car 2 starts from town B. They move towaRD each other with constant speeds: Car 1 at 50 mph, car 2 at 40mph.

A. How far apart are the two cars if car 2 has travel a time triangle t (change of t)

B. Where do the cars meet? If you use an equation intrpret the expression on each side and tell why you set them equal.


QUESTION 2

Two objects start at 30km apart at the same time and move toward each other with uniform motion. Object A has a speed of 6.3 km/h and object B has a speed of 4.5 km/h. When will they meet.


QUESTION 3
on a straight inclined track, a ball rolling uphill takes 1.3 seconds to slow down from a velocity of 135cm/s to a velocity of 65 cm/s. On the same track how much time would it take a rolling ball with velocity 85 cm/s to reach a velocity of 0? Explain your reasoning. Do not use algebra

Answer 1

Step 1 is to set up a frame of reference. Set town at at the origin of the x axis and town b at x=200 miles

The position of car1 in this frame of reference is
x1(t)=50*t (t in hours, x in miles)
The position of car 2 in this frame of reference is

x2(t)=200-40*t (t in hours, x in miles)

to find out how far apart the cars are at any given time take
x2(t)-x1(t) if negative, take the magnitude
note that when t=0 the distance is 200 miles
at t=dt the distance is
200-40*dt-50*dt
or
200-90dt
note their relative speed is 90 mph

for when they meet it is when x1(t)=x2(t)
or
200-40*t=50*t
solve for t
t=200/90
the position is
x=50*200/90
and
x=200-40*200/90
x=111 miles

Question 2 is more of the same
x1(t)=6.3*t
x2(t)=30-4.5*t
when x1=x2 they meet, solve for t
t=30/10.8 hours

Question 3 is related to acceleration, a
v(t)=v0-a*t
65=135-a*1.3
solve for a in cm/s^2
a=70/1.3
now assume the same acceleration
0=85-70*t/1.3
or 85*1.3/70 seconds
1.58 seconds

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Answer 2

Part A is so basic that if you can't rattle off the formula to use, then you didn't understand the material on kinetic friction. Part B is interesting because the angle has two effects: The deflection from the horizontal make more pull necessary, but the addition of a vertical component reduces the frictional force. Which will win? Let T be the tension in the rope...the pulling force required. The x component is T cos θ, where θ=30 is the angle above horizontal, and the vertical component is T sin θ. For zero acceleration, you need: (T cos θ) + F = 0 .... no net horizontal force (T sin θ) + N - mg = 0 .... no net vertical force .... where N is the upward normal force of snow on sled, and F is the frictional force, given by:F = -μN The second force equation can be rewritten: N = mg - T sin θ ...and substituted for N above: F = -μN = =μ(mg - T sin θ) (T cos θ) + F = (T cos θ) - μN = (T cos θ) - μ(mg - T sin θ) = 0 (T cos θ) - μmg + μ(T sin θ) = 0 T(cos θ + μ sin θ) = μmg T = μmg / (cos θ + μ sin θ) That's it. The rest is calculation. The numerator (the answer to part A, by the way) is 0.050 * (17 kg) * (9.81 m/s²) = 8.34 N The denominator is cos 30 + 0.050*(sin 30) = sqrt(3)/2 + 0.050*(1/2) = 0.891 The tension is then T = 8.34 N / 0.891 = 9.36 N. Edit: With only 2 digits in the givens, I probably should have used 9.8 instead of 9.81. That gives 8.3 N of the other answer, and by a rounding difference part B's answer is 9.3 N. Take your pick.