Stuck On Uniform Velocity - Applied Maths

robert12345

Hey guys,

Can you help me with this question please?

A lift starts from rest and travels with constant acceleration 4m/s^2. It then travels with uniform speed and finally comes to rest with constant retardation 4m/s^2. The total distance travelled is d and the total time taken is t.

(i) Show that the time spent travelling at a constant speed is √t^2 -d

Reference: Applied Mathematics 2nd Edition Oliver Murphy Page 35 Question 8

Brendan1234

First off make a velocity-time graph, which is actually the first part of that question in the book.

Now, the area under a velocity-time graph is equal to the total distance covered. Because the lift accelerates and decelerates at the same rate, it takes the same amount of time for each. Call this time t(1). The distance covered while accelerating = 1/2 x t(1) x v, where v is the maximum velocity reached. Since the lift covers the same distance while accelerating and decelerating, the sum of these = t(1) x v. Let t(2) be the time spent moving at a constant speed. The distance covered during this time = v x t(2)
So the equation for the total distance:

(t(1) x v) + (t(2) x v) = d

Now, use the equation, a = (v-u)/t, where a=acceleration, t= t(1) and u= the initial speed, which in this case is 0
So the equation becomes:

4=v/t(1)
v=t(1) x 4

Also:

(2 x t(1)) + t(2) = t, the total time
t(1) = (t-t(2))/2

Substitute the second two equations into the first one and you get the answer.
Sorry if it's hard to follow, I'm not great at explaining things :P