Question

A train, travelling along a straight horizontal track, has a steady speed of 18m/s as it passes the point A. Fifteen seconds later, it begins to slow down at a uniform rate for 30 s until its speed is 10m/s. The train then increases its speed for 45 s until it reaches a speed of 20 m/s as it passes the point B. a) b) Calculate the acceleration of the train just before it reaches B.

Answer 1

Draw a sketch of the journey traveled by the train:
i)first part: travels through point A with a speed of 18m/s
ii)second part: the train leaves point A to a point b and after  15 seconds later slows down to 10m/s in 30s to a point b.
iii)leaves point b and passes through point B after 45s with a speed of 20 m/s.

The first part of the question (a) is missing, but to answer part (b):
Acceleration of the train before it reaches B:
Use kinematic equation-----v2= u2 + 2as

SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-timev=u + atv2 = u2 + 2ass=1/2(u+v)ts=ut + 1/2at2

Before the train left for part it had the initial velocity of 10 m/s.(u)
the train passed through point B with a velocity of 20 m/s.(final velocity v)
The train traveled from point b to point B in 45s(t)

There acceleration a:
v2= u2+2as
20^2= 10^2 + 2 x a x 45
400= 100 + 90a              90a= 300
                                            a=300/90
                                              =3.33m/s
Therefore the acceleration was 3.33 m/s
 

Answer 2

Answer:

Explanation:

Ok