Question

A train is travelling at a speed of 90 km per hour breaks are applied so as to produce a uniform acceleration of _i.5m per sec square. find how far the train will go before it is in rest.

Answer 1

Answer:

208.333... metres

Explanation:

Given:

• Initial velocity of the train is given as 90 km/h
• Acceleration of the train = a = -1.5 m/s²
• Final velocity of the train = v = 0 m/s

To find:

• Distance travelled by the train before coming to rest

u = 90 km/h = $90 \times \frac{5}{18} = 25 \ m/s$

Using third equation of motion:

V²-U²=2as

0²-25²=2×-1.5×s

0-625= -3s

-625 = -3s

s = $\frac{625}{3}$

s = 208.33... metres

The train will travel a distance of 208.33... metres