Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (I) the acceleration and (ii) the distance traveled by train while it attained this velocity.

Answer 1

Step 1:Given data:

Initial velocity, u = 0.

The velocity of the train, v = 72 km/h.

Time of travel, t = 5 min.

Acceleration is uniform.

Step 2:Finding the acceleration:

Let, the acceleration of the train is a.

$v=72km/h=\frac{72\times 1000}{3600}=20m/s.$

From Newton's first law of motion,

$v=u+at$.

Where v and u are final and initial velocities respectively, and a is a constant acceleration and t = required time.

Now,

$$\begin{gathered} v=u+at. \\ ora=\frac{v-u}{t}. \\ ora=\frac{20-0}{5\times 60}=\frac{20}{300}=\frac{1}{15}m/s^2. \end{gathered}$$

Therefore the acceleration of the train is $\frac{1}{15}m/s^2.$

Step 3:Finding the distance:

Let, the distance is S.

We know that,

From the second equation of motion,

$v^2-u^2=2as.$

where v and u are the final and initial velocities respectively, a = acceleration, s = distance travelled.

So,

$$\begin{gathered} v^2-u^2=2as. \\ s=\frac{v^2}{2a}=\frac{{20}^2}{2\times {\displaystyle \frac{1}{15}}}=\frac{400\times 15}{2}=3000m. \end{gathered}$$

Therefore, the distance traveled by train is 3000 m.