Question

a train starts from rest and accelerates uniformly at 100 min-2 for 10 min .It then maintains a constant velocity for 20 minutes.the brakes are then applied,and the train is uniformly retarded.it comes to rest in 5 min.find the maximum velocity retarded and the total distance travelled?

Answer 1

Taking

Acceleration (a) = 100 m/min2

t1 = 10 min

As the train starts from rest, u = 0

Answer 2

The maximum velocity retarded is 1000 m/s

The total distance travelled is 27.5 km

Explanation:

Let ‘u’ be initial velocity.

Let ‘v’ be final velocity.

The final velocity of the train is:

$\bold{v = u + at}$

$\bold{v = 0 + (100 \times 10)}$

v = 1000 m/s

So, this is the maximum velocity of the train.

The distance covered is given by the formula:

$\bold{S = \frac{1}{2}at^2}$

Distance covered during uniform acceleration:

$\bold{S_{1} = \frac{1}{2}at^{2}_{1}}$

$\bold{S_{1} = \frac{1}{2} \times 100 \times 10^{2}}$

$\bold{S_{1} = 5000 \ m = 5 \ km}$

Distance covered during constant velocity:

$\bold{S_{2} = v \times t_{2}}$

$\bold{S_{2} = 1000 \times 20}$

$\bold{S_{2} = 20000 \ m = 20 \ km}$

Distance covered during retardation:

Retardation = $\bold{200 \ m/min^{2}}$

$\bold{S_{3} = \frac{1}{2}at^{2}_{3}}$

$\bold{S_{3} = \frac{1}{2} \times 200 \times 5^{2}}$

$\bold{S_{3} = 2500 \ m = 2.5 \ km}$

Now, the total distance is:

$\bold{S = S_{1} + S_{2} + S_{3}}$

$\bold{S = 5 + 20 + 2.5}$

$\bold{\therefore S = 27.5 \ km}$