A train accelerates from 18 km/h to 72 km/h in 10 sec. The distance travelled by train?

Question

Question

Answer 1

The Correct Answer is 125 m.

Concept:

The average acceleration for a given interval is defined as the velocity change for that particular interval of time.
Unlike acceleration, for a given interval the average acceleration is determined.

\(a=\frac{\Delta v}{\Delta t}\)

Δv = Change in velocity

Δt = The duration of the period

Calculations:

Given that,

The initial velocity u = 18 km/h = 5 m/s

The final velocity v = 72 km/h = 20 m/s

Since the answer must be in m so, we need to convert

18 km/hr converted to 18 × (5/18) = 5m/s

72 km/hr converted to 72 × (5/18) = 20 m/s

Time = 10 sec

To find the distance travelled by car, we need to find acceleration,

\(a=\frac{\Delta v}{\Delta t}=\frac{change\; in\; velocity}{time}=\frac{20-5}{10}=1.5\;ms^{-2}\)

a = 1.5 m/s²

The distance travelled by car, we use the 2^nd equation of motion,

\(s=ut+\frac{1}{2}at^{2}\)

\(s=(5\times 10)+\frac{1}{2}(1.5)\times 10^{2}\)

s = (50 + 75) m

s = 125 m

The distance travelled by car is 125 m.