Question

A train accelerates from 36km/h to 54km/h in 10 sec. Find

a) Acceleration

b) The distance travelled by the train.

Answer 1

Step 1: Given that:

Initial velocity(u) of train = $36km{h}^{-1}$ = $36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$

final velocity(v) of the train = $54km{h}^{-1}$ = $54\times \frac{5}{18}m{s}^{-1}=15m{s}^{-1}$

Time(t) = $10sec$

Step 2: a) Calculation of acceleration:

Using the first equation of motion; $v=u+at$

Where; u = initial velocity, v= final velocity, a = acceleration and t = time

Putting the given values, we get;

$15m{s}^{-1}=10m{s}^{-1}+a\times 10sec$

$10+10a=15$

$10a=15-10$

$10a=5$

$a=5m{s}^{-2}$

Step 3: b) Calculation of distance travelled by train:

Using the second equation of motion; $s=ut+\frac{1}{2}a{t}^2$

Where; s = distance covered by the body.

Putting the values, we get;

$s=10m{s}^{-1}\times 10sec+\frac{1}{2}\times 5m{s}^{-2}\times {\left(10sec\right)}^2$

$s=100+2.5\times 100$

$s=100+250$

$s=350m$

Thus,

a) The acceleration of the train = $5m{s}^{-2}$

b) The distance covered by the train = $350m$