Question

The position of a particle is given by the eq. s=t^3-6t^2+9t.
Find the displacement at t=5 sec.

Answer 1

GiveN :

• s = t³ - 6t² + 9t

To FinD :

• Displacement after 5 sec

SolutioN :

Here, displacement is given by :

$\longrightarrow \sf{s \: = \: t^3 \: - \: 6t^2 \: + \: 9t}$

We have to find the displacement after 5 s. So, simply put t = 5s.

$\implies {\sf{s \: = \: 5^3 \: - \: 6(5)^2 \: + \: 9(5)}} \\ \\ \implies {\sf{s \: = \: 125 \: - \: 6(25) \: + \: 45}} \\ \\ \implies {\sf{s \: = \: 170 \: - \: 150}} \\ \\ \implies {\sf{s \: = \: 20 \: m}} \\ \\ \underline{\sf{\therefore \: Displacement \: after \: 5 \: s \: is \: 20 \: m}}$

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Additional Information :

• Displacement is the shortest path travelled by a body.

• It is a vector quantity.

• It's SI unit is meter (m)

• Displacement = Velocity * Time

Answer 2

Answer:

Given, the position of particle by an equation :-

s = t³ - 6t² + 9t

Here, the equation indicates displacement.

Time given, t = 5 sec.

Hence, s = (5)³ - 6(5)² + 9(5)

s = 125 - 6*25 + 45

s = 125 - 150 + 45

s = 170 - 150

s = 20 m.

Hence, the Displacement = 20 m.

• Displacement is the shortest distance known between the initial position and final position of the particle/object.
• It is a vector quantity.
• Displacement = Velocity*time
• The SI unit = m.
• Sometimes, distance = displacement.
• Velocity can be defined as displacement per unit time.