Question

A particle moves in space such that
x=2t? + 3t+4;
y=t? +4t -1;
z=2 sin it
where x, y, z are measured in meter and t in second. The
acceleration of the particle at t=3s is
(a) 364+2j+k ms.
(b) 36î +29+ tk ms?
(0)36 +29 ms 2​

Answer 1

Answer:

refer to the attachment...........

Answer 2

Therefore the total acceleration of the particle is 36i + 2j m/s². ( Option-c )

Given:

A particle moves in space such that

x = 2t³ + 3t + 4;

y = t² + 4t - 1;

z=2 sin πt

To Find:

The acceleration of the particle at t = 3s.

Solution:

The given question can be solved as shown below.

Given that,

x = 2t³ + 3t + 4;

y = t² + 4t - 1;

z=2 sin πt

Velocity in x-direction = u = dx/dt = d( 2t³ + 3t + 4 )/dt ⇒ u = 6t² + 3

Acceleration in x-direction = a₁ = du/dt = ( 6t² + 3 ) = 12t

Velocity in y-direction = v = dy/dt = d( t² + 4t - 1 )/dt = 2t + 4

Acceleration in y-direction = a₂ = dv/dt = d( 2t + 4 )/dt = 2

Velocity in z-direction = w = dz/dt = d( 2 sin πt )/dt = 2π cos πt

Acceleration in z-direction = a₃ = dw/dt = 2 d cos πt/dt = -2π² sin πt

⇒ Acceleration of the particle in the x-direction at 3 seconds a₁ =12 × 3 = 36 m/s²

⇒ Acceleration of the particle in the y-direction at 3 seconds a₂ =2 m/s²

⇒ Acceleration of the particle in the z-direction at 3 seconds a₃ = -2π² sin 3π m/s² = 0 m/s²

Total Acceleration of the particle = a₁i + a₂j + a₃k = 36i + 2j

Therefore the total acceleration of the particle is 36i + 2j m/s².

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