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    PracticeAQA GCSEAQA GCSE Maths Higher Tier Paper 3 CalculatorQuestion 23
    Medium3 marksStructured
    AlgebraHigherAlgebraGraphsExponential Functions

    AQA GCSE · Question 23 · Algebra

    t d 0 5 10 15 20 25 30 35 40 45 50 0 1 2 3 4

    The distance of a particle from a point is d metres after t seconds. The relationship is given by d = a × bᵗ where a and b are constants. Use the graph to work out the values of a and b.

    How to approach this question

    1. The equation is d = a × bᵗ. To find 'a', consider what happens when t = 0. Any number to the power of 0 is 1. 2. Find the value of d on the graph where t = 0 (the y-intercept). This value is 'a'. 3. Now that you have 'a', rewrite the equation with this value. 4. To find 'b', pick another clear point on the graph with integer coordinates (e.g., where t=1, t=2, or t=3). 5. Substitute the t and d values from this point into your equation. 6. Solve the resulting equation for b.

    Full Answer

    1. **Find the value of a.** The constant 'a' represents the initial value, which is the value of d when t = 0. From the graph, when t = 0, the curve intersects the d-axis at d = 5. So, **a = 5**. (Mathematically: d = a × b⁰ => d = a × 1 => d = a) 2. **Find the value of b.** Now we know the equation is d = 5 × bᵗ. We can pick another point from the graph to find b. Let's choose an easy point to read, like (t=2, d=20). Substitute these values into the equation: 20 = 5 × b² Divide by 5: 4 = b² Take the square root: b = 2 (we take the positive root as the graph is increasing). So, **b = 2**. 3. **Check with another point.** Let's check with (t=3, d=40). d = 5 × 2³ = 5 × 8 = 40. This matches the graph, so our values are correct. Final answer: a = 5, b = 2.
    The equation given is for an exponential curve: d = a × bᵗ. Step 1: Find the value of `a`. The constant `a` is the initial value, or the y-intercept, where t=0. Let's substitute t=0 into the equation: d = a × b⁰ Since any number raised to the power of 0 is 1 (b⁰ = 1), the equation becomes: d = a × 1 d = a So, `a` is the value of `d` when t=0. Looking at the graph, the curve crosses the d-axis at 5. Therefore, **a = 5**. Step 2: Find the value of `b`. Now we know the equation is d = 5 × bᵗ. We can find `b` by substituting the coordinates of another point from the curve. Let's choose a point that is easy to read accurately, for example, the point (t=2, d=20). Substitute t=2 and d=20 into the equation: 20 = 5 × b² To solve for b, first divide both sides by 5: 20 / 5 = b² 4 = b² Now, take the square root of both sides: b = √4 b = 2 (We take the positive root because the graph shows growth). Therefore, **b = 2**. The final equation is d = 5 × 2ᵗ. We can check this with another point, (t=3, d=40): 40 = 5 × 2³, which is 40 = 5 × 8. This is correct.

    Common mistakes

    ✗ Reading the points from the graph inaccurately. ✗ Mixing up a and b. ✗ Trying to treat it as a linear equation and finding a gradient. ✗ Making an error when solving for b, e.g., forgetting to take the square root.
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