IGCSE Revision (Differentiation: Displacement, Velocity and ... - Free Printable

It looks like you've uploaded a worksheet titled "Differentiation: Displacement, Velocity & Acceleration" from Maths4Everyone.com, which contains several problems related to calculus in the context of motion. The tasks involve finding velocity (the derivative of displacement) and acceleration (the derivative of velocity or second derivative of displacement).

Let’s go through some of the key problems on this sheet and solve them step-by-step.

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🔷 Understanding the Concepts

- Displacement: $ s(t) $
- Velocity: $ v(t) = \frac{ds}{dt} $
- Acceleration: $ a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} $

We’ll use basic differentiation rules:
- Power Rule: $ \frac{d}{dt}(t^n) = n t^{n-1} $
- Constant Rule: Derivative of constant is 0
- Sum/Difference Rule: Differentiate term by term

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## ✔ Problem A2:
> The displacement of a particle is given by
> $ s = 2t^2 + 4t + 5 $
> Find an expression for the velocity at time $ t $.

Solution:

Differentiate $ s $ with respect to $ t $:

$$
v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 + 4t + 5)
= 2 \cdot 2t + 4 + 0 = 4t + 4
$$

✔ Answer: $ v = 4t + 4 $

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## ✔ Problem A3:
> The displacement of a particle is given by
> $ s = 5t^2 + 3t - 2 $
> Find an expression for the acceleration at time $ t $.

Step 1: Find velocity

$$
v = \frac{ds}{dt} = \frac{d}{dt}(5t^2 + 3t - 2) = 10t + 3
$$

Step 2: Find acceleration

$$
a = \frac{dv}{dt} = \frac{d}{dt}(10t + 3) = 10
$$

✔ Answer: $ a = 10 $ m/s² (constant acceleration)

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## ✔ Problem B1:
> The displacement of a particle is given by
> $ s = t^3 - 2t^2 + 2 $
> Find the velocity when $ t = 2.5 $

Step 1: Find velocity function

$$
v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 2t^2 + 2) = 3t^2 - 4t
$$

Step 2: Plug in $ t = 2.5 $

$$
v(2.5) = 3(2.5)^2 - 4(2.5) = 3(6.25) - 10 = 18.75 - 10 = 8.75
$$

✔ Answer: $ v = 8.75 $ m/s

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## ✔ Problem B3:
> The velocity of a particle is given by
> $ v = 6t^2 - 5t $
> Find the acceleration when $ t = 3 $

Step 1: Find acceleration

$$
a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 5t) = 12t - 5
$$

Step 2: Plug in $ t = 3 $

$$
a(3) = 12(3) - 5 = 36 - 5 = 31
$$

✔ Answer: $ a = 31 $ m/s²

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## ✔ Problem C4:
> The displacement of a particle is given by
> $ s = t^3 - t^2 - t - 1 $
> Find the time at which the acceleration of the particle is $ 15 $ m/s².

Step 1: Find velocity

$$
v = \frac{ds}{dt} = 3t^2 - 2t - 1
$$

Step 2: Find acceleration

$$
a = \frac{dv}{dt} = 6t - 2
$$

Step 3: Set $ a = 15 $

$$
6t - 2 = 15 \\
6t = 17 \\
t = \frac{17}{6} \approx 2.83 \text{ seconds}
$$

✔ Answer: $ t = \frac{17}{6} $ s

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## ✔ Problem D4:
> The velocity of a particle is given by
> $ v = 4t^2 - 3t $
> Find a possible expression for the displacement at time $ t $.

Solution:

We integrate velocity to get displacement:

$$
s = \int v \, dt = \int (4t^2 - 3t) \, dt = \frac{4}{3}t^3 - \frac{3}{2}t^2 + C
$$

Since no initial condition is given, we leave $ C $ as a constant.

✔ Answer: $ s = \frac{4}{3}t^3 - \frac{3}{2}t^2 + C $

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📌 Summary of Key Rules:

| Quantity | Formula |
|----------------|----------------------------------|
| Velocity | $ v = \frac{ds}{dt} $ |
| Acceleration | $ a = \frac{dv}{dt} = \frac{d^2s}{dt^2} $ |
| Displacement | $ s = \int v \, dt $ |

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