Question

Please answer this question ​

Answer 1

Hey friend!

Here is your answer:

Given,

→ A particle is moving along the x- axis

→ The position of the particle at any instant is given by x = 20 + 0.1t²

→ x is measured in metres and time (t) is measured in second.

To Find:

i) Average Acceleration of the particle between t = 2s to t = 3s

ii) Show that the acceleration of the particle is constant.

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We are given that position is a function of time where

x = 20 + 0.1t²

Now,

Method 1 [below grade 11 solution]

For t = 1s

x = 20 + 0.1 × (1)²

⇒ x = 20 + 0.1 × 1

⇒ x = 20.1 m

For t = 2s,

x = 20 + (0.1) × 2²

⇒ x = 20 + 0.1 × 4

⇒ x = 20 + 0.4

⇒ x = 20.4 m  

And for t = 3s,

x = 20 + 0.1 × (3)²

⇒ x = 20 + 0.1 × 9

⇒ x = 20 + 0.9

⇒ x = 20.9 m

Distance covered in between t=1s and t=2s is 20.4 - 20.1 = 0.3 m

Distance covered in between t=2s and t=3s is 20.9 - 20.4 = 0.5 m

To get the Speed is distance ÷ time

Instantaneous Speed at t = 2 s, = (0.3) ÷ (2 - 1) = 0.3 m/s

Instantaneous Speed at t = 3 s, = (0.5) ÷ (3 - 2) = 0.5 m/s

Average acceleration = (final velocity - initial velocity)/change in time

                                 = (0.5 - 0.3) ÷ (3 - 2)

                                 = (0.2) ÷ 1

                                 = 0.2 m/s²

Now,

we know s = ut + ¹/₂ at²   where s = s₁ - s₀

and if we compare this with the equation x = 20 + 0.1 t²

here s₀ = 20 m

we get: 0.1 t² = ut + ¹/₂ at²

but [1] exponent power is not in LHS as there is ut In RHS

⇒ No involvement of initial velocity,u

⇒ 0.1 t² = ¹/₂ a t²

⇒ 0.1 × 2 = a

⇒ a = 0.2

⇒ a is independent of t

⇒ acceleration is constant

Method 2 [ 11 th grade and above solution]

Let x = 20 + 0.1t²

Differentiation formulas used here:

1. $\frac{dy}{dx} = 0$ when y = constant
2. $\frac{dy}{dx} = nx^{n - 1}$ when y = xⁿ⁻ ¹

Now,

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

             = $\frac{d}{dt}(20 + 0.1t^2)$

             = 0.2t m/s

Acceleration = $\frac{d^2}{dt^2} x$

                    = $\frac{d}{dt} (0.2t)$

                    = 0.2 m/s²

∴ Average acceleration = 0.2 m/s²

∵ Acceleration is shown as an arbitrary constant here independent of time, t

⇒ Acceleration is constant throughout the motion

I hope it helps you and if you appreciate my answer, please respond with a thanks