Question

The displacement x of a particle at a time t is given by x = 160t -16t², show that its velocity at t = 1 and t = 9 are equal in magnitude but opposite in direction.

Answer 1

sorry i didnt get your question

Answer 2

Answer:

At, t = 1

v = 128

At, t = 9

v = -128

Step-by-step explanation:

In the given question,

The displacement of a particle is given by,

$x =160t-16t^{2}$

So,

As we know that the Velocity(v) is given by the derivative of the displacement(x), w.r.t time (t),

So,

$v=\frac{dx}{dt}$

Now,

$v=\frac{d(160t-16t^{2})}{dt}=160-32t$

So,

At, t = 1

Velocity is given by,

$v=160-32t\\v=160-32(1)\\v=128$

Also,

At, t = 9

Velocity is given by,

$v=160-32(9)\\v=160-288\\v=-128$

Therefore, we can see that the velocity at t = 1 and t = 9 are both same in magnitude but opposite in the direction as one is positive and other is negative.