Question

Each of four particles move along an x axis. Their coordinates (in meters) as functions of time(in seconds) are given byparticle 1: x(t) = 3.5 ? 2.7t3particle 2: x(t) = 3.5 +2.7t3particle 3: x(t) = 3.5 +2.7t2particle 4: x(t) = 3.5 ? 3.4t ? 2.7t2Which of these particles have constant acceleration?A. All fourB. Only 1 and 2C. Only 2 and 3D. Only 3 and 4E. None of them

Answer 1

Answer:

D OPTION

Explanation:

coordinates (in meters) as functions of time(in seconds) are given by the equation

1 ) x(t) = 3.5 +2.7t⁴

differentiating on both sides

dx / dt = 2.7 x 4 t³

Again differentiating

d²x / dt² = 2.7x 4 x 3 t²

= 32.4 t²

d²x / dt²  is acceleration so this  acceleration depends upon time .

2 ) x(t) = 3.5 +2.7t³

dx/dt = 2.7 x 3 t²

again differentiating

d²x/dt² = 2.7 x 3 x 2 t

= 16.2 t

This acceleration also depends upon time

3 ) x(t) = 3.5 +2.7t²

dx / dt = 2.7 x 2 t

again differentiating

d²x / dt² = 2.7 x 2 = 5.4

This acceleration is independent of time so it is constant.

4 )

x(t) = 3.5 + 3.4t + 2.7t²

differentiating both sides

dx / dt = 3.4 + 2.7 t

again differentiating

d²x / dt² =  2.7

This acceleration also is constant .

so D OPTION is corect

Answer 2

Answer:

D. Only particle 3 and 4

Explanation:

The acceleration of the particle can be found out easily by taking the derivative of each function, twice.

So, when we try to find the second derivatives of all the function, we get to know that only particle 3 and 4 will have a constant value. All the other particles will still be a function of "t". Thus 'D' is the correct answer.

The second derivatives will be as follows:

Particle 1: a(t) = 16.2 t

Particle 2: a(t) = 16.2 t

Particle 3: a(t) = 5.4

Particle 4: a(t) = 5.4