Need some help seeing how to do these problems more efficiently so I\'m looking

Question

Need some help seeing how to do these problems more efficiently so I'm looking for some other ways to solve.

PHY 251: Master Set: Ch. 02 One-Dimensional Motion Deieitions and Basics 1. Under what condition will a motion problem require the use of cakculues to solve? 2. Which of the following unuld be an indication ofan (yet undergoing an ? f) not moving a) changing direction cmoving aroand a curve b) jerk s not d moving at a constant speed slowing down e moving at a constant velocity h) speeding up 3. A box is slid akong a level frictionless oor to the right and then up a frictionless ramp. The box reaches a certain point on the ramp when it naturailly tums around, slides back down the ramp to the level floor until it hits a wall. Assame that the positicon-0 just aher the transition from the floor to the ramp. The surface has friction trom the starting point to the bottom of the ramp; everywhere else is frictionless, Let the + direction be lo the right and up the ramp For ease of conveying the answer, feel free to use the following notation: A - the moment when the probkem starts. The box is already moving to the right B- the moment just after the box finishes the transition froim floor to ramp C- the moment at which the box turns around D the moment just before the box starts the transition from ramp to floor E the moment at which the box reaches its starting point but is traveling in the opposite direction F the moment just before the box touches the wall. For cach of the following situations, write down the parts (using the [I.0 sotation) of the boa's movement from A to F with the situations, below. If a situation can not hippen, then write "N/'A a) + Accelersion, a0 b) 0 Acceleration, - c) -Acceleration, ust0 d) + Velocity, 0 e) 0 Velocity, -0 ) -Velocity g) + Displacement, i) -Displacement, 0 k) 0 Position, l) -Position, x

Explanation / Answer

Hi,

you have not mentioned which of the problems to solve, so i am solving the first 4 problems according to our protocol.

1)The question is not correct. You can solve every motion problem with the help of calculus. There are no "conditions" whether it would require calculus or not. Calculus is a mathematical method and it can be used anywhere in any way you want.for eg. if i ask what is the distance travelled with a constant speed of 5m/s in 10s then you may used speed = distance/time formula or use v=ds/dt then ds = vdt and integrate both sides with limits to get the answer.

2)acceleration is non-zero means velocity is changing wrt time.

a) by changing direction velocity changes, thus a is non zero.

b) here, sometimes a would be zero, sometimes not. for eg take an acceration vs time graph to be concave up graph with a local minimum at a=0.da/dt is non-zero but at the minima a is zero and at other points non zero. So, b) is not an indication for a to be non-zero.

c)moving around a curve changes direction, so velocity changes, hence a is non zero.

d) speed does not account for an objects direction. direction might not change or might not, so a might be no-zero or might be.

e)constant velocity means acceleration is zero as a = dv/dt.

f) not moving can also mean that the object is instantaniously at rest, it might have an acceleration at that time so a can in somecases be non-zero. But it would not indicate everytime that an object is not moving and it has a non zero acceleration.

g)slowing down means acceleration is negative, a is non zero.

h)speeding up means a is positive.

3)

the problem says there is friction from starting point A to bottom of ramp B. So AB has friction, rest does not.

friction would provide a constant deceleration.So the velocity vs time graph would decrease at a constant rate till the object reaches B. going up and down the ramp, the total mechanical energy would remain constant, with Potential energy increasing till C and then decreasing. So kinetic energy and so the velocity would first decrease and then increase back to the value it had at B. After moving through DE with a constant deceleration due to friction its velocity would decrease till E and then remain constant till F and then finally zero at F.

therefore,

a)CD

b)EF

c)AC and BA

d)AC

e)C & F

f)CF

g)[AE)

h)[AE]

i)EF

j)AC

k)E

l)EF

4)

a)Since the acceleration (-5m/s2)is constant we can directly use the motion formula v2=u2 +2*a*s

where v and u are final and initial velocity and a and s are accelertion and displacement.

v= 14.526m/s

b) using v-u =a*t

v=0

therefore, t= 3.8s

c)use the formula s=u*t+1/2*a*t^2

directly put the values to get the answer -8m.

that will be the displacement.

since after 3.8s the velocity becomes negative, the object has thus started moving in the negative direction.

in calculation of distance(that is the length of the path) we will have to calculate the individual displacements first till t=3.8s and then from 3.8s till 8s.distance travelled in first 3.8s = 36.1m

then it comes back to origin(36.1m again travelled backwards) and then 8m in the neg direction from origin(as we can see from the value of displacement).

total distance travelled=36.1+36.1+8=80.2m

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.