M1

pg. 51, Q6)
A train starts from rest and moves with uniform acceleration for 5 minutes. It then maintains a constant speed for 20 minutes before being brought to rest by a uniform retardation of magnitude twice that of the acceleration. Sketch a speed-time graph of the motion and find the time it takes the train to stop. If the train travels 4.5km while accelerating, show that the accleration is 1/ 1 0m/s²and find the total distance travelled.

Reply 1

bob_54321

the graph is a straight diagonal line up, then a flat line for 20 mins, and then a striaght diagonal line down at twice the (negative) gradient.

u = 0, s=4500, t= 5*60=300

s = ut +1/2 at^2
4500 = 1/2 a *90000
9000 = 90000a
a = 0.1

v = u + at
v = 0.1*300
v = 30ms^-1

20*60*30=12000m (for constant speed bit)

for final bit, if u =30, v=0, a=-0.2
v^2 = u^2 + 2as
900 = 0.4s
s = 2250

total distance = 2250 + 4500 + 12000 = 18750 m = 18.75km i think

Reply 2

turtle2

OP

7

The answers are 150s and 42.8km

Reply 3

bob_54321

sorry, silly silly mental arithmetic error with the distances, 20*60*30 = 36000 (not 12000!!!), so total distance = 2250 + 4500 + 36000 = 42750m = 42.75km

and i missed the first part

use v = u + at
0 = 30 - 0.2t
t = 150s

Reply 4

turtle2

OP

7

ohh ok, it makes sense! thanks!

Reply 5

turtle2

OP

7

use v = u + at
0 = 30 - 0.2t
t = 150s

I dont really understand where the -0.2 for a comes from now...

Reply 6

bob_54321

ok, well, you know that the acceleration is 0.1 ms^-2 originally, and it says that there is "a uniform retardation of magnitude twice that of the acceleration". so, 2 * 0.1 = 0.2. that is twice the magnitude. since it is a retardation (negative acceleration), then that is where the minus comes from.

Reply 7

turtle2

OP

7

ohh yeah, i forgot that it's twice the acceleration...oops :redface:

thanks!

M1

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