# Kinematics: The Motion of Particles

• October 9th 2009, 12:02 PM
BabyMilo
Kinematics: The Motion of Particles
A train takes 5 minutes to cover 3km between stations P and Q. Starting from rest at P, it accelerates at a constant rate to a speed of 40km/hr and maintains this speed for a while and then is brought uniformly to rest at Q. If the braking takes three times as long as the accelerate, find the time taken for the train to accelerate.

I worked out that 3000=0*t+0.5at^2+(100/9)*t+0.5a*3t^2+100/9*16t applies to the answer. and t=15s, a= (20/27)ms^-2

but how do you work t and a out?

Million thanks!
• October 9th 2009, 11:19 PM
Hello BabyMilo
Quote:

Originally Posted by BabyMilo
A train takes 5 minutes to cover 3km between stations P and Q. Starting from rest at P, it accelerates at a constant rate to a speed of 40km/hr and maintains this speed for a while and then is brought uniformly to rest at Q. If the braking takes three times as long as the accelerate, find the time taken for the train to accelerate.

I worked out that 3000=0*t+0.5at^2+(100/9)*t+0.5a*3t^2+100/9*16t applies to the answer. and t=15s, a= (20/27)ms^-2

but how do you work t and a out?

Million thanks!

Since the speed we're given is measured in km/hr, and the distance in km, let's suppose that the train accelerates for $t$ hr. So it brakes for $3t$ hr and travels at constant speed for $(\tfrac{1}{12}-4t)$ hr, since $5$ min $= \tfrac{1}{12}$ hr.

Draw a velocity-time graph, which is in the form of a trapezium with horizontal, parallel sides of length $\tfrac{1}{12}$ (along the t-axis) and $(\tfrac{1}{12}-4t)$ along the line where $v = 40$.

Now the area under this graph represents the distance travelled. So:

$\tfrac12$ (sum of parallel sides) x (perpendicular distance between them) $=\tfrac12(\tfrac{1}{12}+\tfrac{1}{12}-4t)40=3$

$\Rightarrow 20(\tfrac{1}{6}-4t)=3$

$\Rightarrow 4t=\frac{1}{6}-\frac{3}{20}=\frac{1}{60}$

$\Rightarrow t = \frac{1}{240}$ hr.

So the acceleration time $= \tfrac14$ min $= 15$ seconds.