Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by

1) particle 1: x (t) = 3.5 – 2.7 t^{3}

2) particle 2: x (t) = 3.5 + 2.7 t^{3}

3) particle 3: x (t) = 3.5 – 2.7 t^{2}

4) particle 4: x (t) = 3.5 – 3.4t - 2.7 t^{2}

Which of these particles have constant acceleration?

This question was previously asked in

ISRO URSC Technical Assistant Mechanical held on 24/03/2019

Option 4 : Only (3) and (4)

ISRO VSSC Technical Assistant Mechanical held on 09/06/2019

2302

80 Questions
320 Marks
120 Mins

**Concept:**

If X = f(t) defines a position vector of a particle then the first derivative represents the velocity of the particle and the second derivative represent the acceleration of the particle.

Now If, \(\frac{d^2x}{dt^2}\ne f(t)\) then it represents a **constant acceleration.**

**Calculation:**

**Given:**

particle 1: x(t) = 3.5 - 2.7 t^{3}, particle 2: x(t) = 3.5 + 2.7 t3, particle 3: x(t) = 3.5 - 2.7 t3, particle 4: x(t) = 3.5 - 2.7 t3

The second derivative of Particle 1:

\(\frac{{{d^2}}}{{d{t^2}}}{x(t){}} = \frac{{{d^2}}}{{d{t^2}}}(3.5 - 2.7 t^3)\)

\(\frac{{{d^2}}}{{d{t^2}}}{{x(t)}} = \frac{d}{{dt}}\left( {0 - 2.7 \times 3\;{t^2}} \right)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x(t){}} = - 16.2\;t\)

**which is a function of time 't' hence acceleration is not constant.**

The second derivative of Particle 2:

\(\frac{{{d^2}}}{{d{t^2}}}{x{(t)}} = \frac{{{d^2}}}{{d{t^2}}}(3.5 + 2.7 t^3)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = \frac{d}{{dt}}\left( {0 + 2.7 \times 3\;{t^2}} \right)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = 16.2\;t\)

which is a function of time 't' hence acceleration is not constant.

The second derivative of Particle 3:

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = \frac{{{d^2}}}{{d{t^2}}}(3.5 - 2.7 t^2)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = \frac{d}{{dt}}\left( {0 - 2.7 \times 3\;{t}} \right)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = - 8.1\)

which is not a function of time 't' hence acceleration is constant.

The second derivative of Particle 4:

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = \frac{{{d^2}}}{{d{t^2}}}(3.5 - 3.4t - 2.7{t^2})\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = \frac{d}{{dt}}\left( {0 - 3.4 - 2.7 \times 2\;{t^{}}} \right)\)

\(\frac{{{d^2}}}{{d{t^2}}}{x{\left( t \right)}} = -5.4\)

which is not a function of time 't' hence acceleration is constant.

Hence only particle 3 and particle 4 have Constant acceleration.