Question

a train was moving at the rate of 54 km/hr when the brakes were applied.It came to rest within a distance of 225m.calculate the retardation produced in the train. (hint: retardation = acceleration & use here third law of motion)

Answer 1

Given:-

Initial velocity(u)=54km/h

=>54×5/18

=>3×5

=>15m/s

Final velocity(v)=0 (as it finally comes to rest after brakes applied)

Distance(s)=225m

To find:-

Retardation(or negative acceleration)

Solution:-

Let the retardation(negative acceleration)be a

Here,we will use Newton's third equation of motion i.e.

$v ^{2} - {u}^{2} = 2as$

0-(15)²= 2×a×225

-225=450a

a= -225/450

a= -0.5m/s²

Since the acceleration is -0.5m/s²,thus retardation is 0.5m/s²

Answer 2

Answer:

Explanation:

u= 54km/hr = 15m/s

v= 0km/hr = 0m/s

s= 225m

From third law of motion

v^2-u^2 =2as

a = (0^2- 15^2)/2 *225

a = -1/2 = -0.5m/s^2

Since retardation is negative of acceleration

Therefore retardation = 0.5m/s^2