Question

A train starts at rest, accelerates with constant acceleration a for 5 minutes, then travels at constant speed for another 5 minutes, and then decelerates with –a. Suppose it travels a distance of 10km in all. Find a.

Answer 1

Answer:

The value of acceleration equals $0.0556m/s^{2}$

Explanation:

The  distances covered in each of the three phases are calculated as under

1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)

Using second equation of kinematics

$s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}$

2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as

$v=u+at\\\\v=0+a\times 300\\\\v=300a$

Thus distance traveled equals

$s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a$

3)

The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as

$v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}$

Now the sum of $s_{1}+s_{2}+s_{3}$ equals 10 kilometers or 10000 meters.

Thus we get

$0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}$