Question

The position vector of a particle is given by r=2ti+3t^2j-5k.The speed of the particle at time=2s is

Answer 1

V=2i+6tj-5k

At t=2,

V=2i+6(2)j-5k

V=2i+12j-5k

V=✓2²+12²+(-5)²

=✓173

=13.15m/s.

Answer 2

Concept:

The speed of an object can be defined as the time derivative of the position of the object.

Given:

Position vector of a particle,  r = 2ti + 3t²j -5k.

Time, t = 2 sec

Find:

The speed of the particle at 2 seconds.

Solution:

As the velocity can be defined as the derivative of the position vector.

v = dr/dt

v = d(2ti + 3t²j -5k)/dt

v = 2i + 6t j

At the time, t = 2 seconds,

v = 2i + 6t j

v = 2i + 6(2) j = 2i + 12 j

The magnitude of the velocity is the speed,

|v| = s = √(2²+12²)

s = √(4+144)

s = √148 m/s

Hence, the speed of the particle at time t = 2 seconds is √148 m/s.

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