PHY 101 Assignment Question and Answers

PHY 101 assignments Update

1. The displacement functions of the particles is given by S=3t²-5t + 4 where x – displacement t- time.
   I. Determine the velocity of the particles of 7 sec after the motion started.
   Ii. Determine the initial velocity
2. The motion of a particle is given by S=T³ + 6t
   Find the displacement of the particles at T=2secs.

Find the acceleration of the particles at t=4sec

Solution

1. The displacement function of the particle is given by S = 3t² – 5t + 4, where S is the displacement and t is the time.

i. Determine the velocity of the particle 7 seconds after the motion started.

To find the velocity, we need to differentiate the displacement function with respect to time (t), since velocity is the rate of change of displacement.

Displacement:

S = 3t² – 5t + 4

Differentiating with respect to t:

v = dS/dt = 6t – 5

Now, substitute t = 7 seconds:

v = 6(7) – 5 = 42 – 5 = 37 m/s

So, the velocity at t = 7 seconds is 37 m/s.

ii. Determine the initial velocity.

The initial velocity is the velocity at t = 0.

Using the velocity function we derived:

v = 6t – 5

Substitute t = 0:

v = 6(0) – 5 = -5 m/s

So, the initial velocity is -5 m/s.

2. The motion of a particle is given by S = t³ + 6t.

i. Find the displacement of the particle at t = 2 seconds.

To find the displacement, we substitute t = 2 into the displacement equation:

S = t³ + 6t

S = (2)³ + 6(2)

S = 8 + 12

S = 20 meters

So, the displacement at t = 2 seconds is 20 meters.

ii. Find the acceleration of the particle at t = 4 seconds.

Acceleration is the rate of change of velocity, and velocity is the derivative of displacement. So we first find the velocity and then the acceleration.

Displacement:

S = t³ + 6t

Differentiating with respect to t to get velocity:

v = dS/dt = 3t² + 6

Now, differentiate velocity to get acceleration:

a = dv/dt = 6t

Substitute t = 4 seconds:

a = 6(4) = 24 m/s²

So, the acceleration at t = 4 seconds is 24 m/s²..