Find total time taken?

Question

The speed of a train increases at a constant rate α = 1 m/s^2 from 0 to v = 10 m/s and then remains constant for a time interval and finally decreases to zero at a constant rate β = 2m/s^2. If l = 100 m is the total distance covered then total time taken is?

A) 17.5 s

B) 2.0 s

C) 13.2 s

D) None of these

Answer 1

take it in steps. First is a period of acceleration:

v = at, t = 10 seconds

In this time the train will cover: distance = 1/2 at^2 = 50m

There is a deceleration period that, by the same logic, takes 5 seconds and covers 1/2 * 2 * 5^2 = 25m.

That leaves 25m that was travelled at a rate of 10 m/s, which take 2.5s.

Add the three periods (in order, 10 + 2.5 + 5) to get 17.5 s (A).

Answer 2

i dunno =/

Answer 3

In the first period (speed of train increases)

vf1 = v01 + a1 t1

10 = 0 + t1

t1 =10 s

average speed = (v01+vf1)/2 = 10/2 =5 m

x1=vt1=5*10 = 50 m

in the third period (the speed of train decreases)

vf3 = v03 + a3 t3

0 = 10 + (-2*t3)

2*t3 = 10

t3 = 5 s

average speed = (v02+vf2)/2 = 10/2 =5 m

x3 = vt3 = 5*5 =25 m

So in the second period (constant speed) distance crossed = 100 -(x1+x3) = 100 -75 =25 m

t2 = x2/v2 =25/10 =2,5 s

So total time taken = t1 + t2 + t3 = 10 + 5 + 2,5 = 17,5 s

Thank you very much

very interesting question