Question

A particle of mass 3 kg moving in a straight line decelerates uniformly from a speed of 40 m/s to 20 m/s in a distance of 300 m. Before it comes to rest, it will travel a further distance pf

A. 10 m B. 20 m C. 50 m D. 100 m

Answer 1

v²= u² + 2 a * s
a= (v²-u²)/2*s
  = (400-1600)/(2*300)
  =-1200/600
  = -2
0= 20² - 2*2*s
s=400/4
  =100 (answer)

Answer 2

Answer:

100 m

Explanation:

It is stated that a particle of mass 3 Kg is moving in a straight line.

It then de-accelerates uniformly from a speed of 40 m/s to 20 m/s while covering 300 m during it.

It then finally comes to a halt.

The motion for this particle can be discussed in two parts.

Using the 3rd Equation of Motion:-

$v^{2}-u^{2}=2as \\ \\(20)^{2}-(40)^{2}=2\times a\times300 \\ \\400-1600= 600a \\ \\a = \dfrac{-1200}{600} \\ \\a = -2 \ m/s^{2}$

The negative sign for acceleration shows the retardation of the particle.

For the rest of the motion:-

20 m/s will act as the initial velocity as the particles finally comes to rest.

$v^{2}-u^{2}=2as \\ \\(0)^{2}-(20)^{2}=2\times (-2)\times S \\ \\-400= -4S \\ \\a = \dfrac{-400}{-4} \\ \\S=100 \ m$

This means that the particle will travel another 100 m before coming to rest.