Newton's laws question.?

Question

A cage carrying workers down a mine shaft completes the 675m descent in 45secs. During the first quarter of its motion, the cage is uniformly accelerated, while in the last quarter it is uniformly retarded, the acceleration and retardation being eqal in magnitude. Calculate the uniform speed of the cage in the central portion of the descent.

I couldn't get this quesion out. Book says answer is 22.5m/s

Answer 1

Use kinematical eqns i think its wrong

as taking average velocity

675/45=15m/s

Answer 2

I think it would be better for you to make a velocity time graph for this so that you better understand what's going on.

for time t=0sec to t=11.25sec the graph will constitute a straight line with +ve slope (this means a constant accln)

Initial velocity would be zero. so it'll pass through origin,

for time t=11.25 to time 33.75 the graph will constitute a straight line(parallel to time axis) which means a constant velocity motion.

for time t=33.75 to t=45 sec the graph will have a negative slope line and the slope will be same as the first quarter line you had drawn since retardation is same magnitude as accln.

Now you can calculate the area under this graph and equate it to your displacement that is 675m

Area = area of triangle in first quarter+ area of rectangle + area of triangle in last quarter which is

(11.25* V)+(22.5*V) = 675

solve this to get your answer

Answer 3

If the acceleration and deceleration are equal, then we know that the distance travelled during this time must also be equal.

We also know that each of these periods took 1/4 of the time, or 45/4 = 11.25 seconds.

Using the distance formulas, we have d=Vit + 0.5at^2 and we also have d=Vt.

Lets call d1 the distance during acceleration, d2 the distance during constant speed, and d3 the distance during decelerartion.  Also, t1 = time during accel, t2 time during constant speedm and t3 is time during decel.

Knowing d1=d3 we get the following:

2(Vit1+0.5at1^2) + V(t2) = 675

We know that the initial velocity, Vi is 0 as the cage starts from rest.  We also know that the final velocity during acceleration is would V = at1 or a = V/t1.  Subbing in we get the following:

2(0.5Vt1) + Vt2 = 675       Note: t2 = 2t1

3Vt1 = 675 where t = 11.25 s.

33.75V = 675 or V = 20 m/s

I don't get 22.5 m/s, but if you back calculate using 22.5 m/s in the formulas above the total distance travelled 759.375 m... so I think it may be incorrect.