1. **State the problem:** Given the speed-time graph of a car, find: a) Acceleration during first 8 seconds. b) Distance covered in first 14 seconds. c) Average speed over 7 seconds. d) Solve equation $2^{x+3} = \frac{1}{16}$. e) Find indefinite integral $\int (6x^2 - 2x + 7) \, dx$. 2. **Part a: Acceleration in first 8 seconds** - The car starts at 0 m/s at $t=0$ and reaches 9 m/s at $t=8$ s. - Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{9 - 0}{8 - 0} = \frac{9}{8} = 1.125$ m/s$^2$. 3. **Part b: Distance covered in first 14 seconds** - Distance = area under speed-time graph from $t=0$ to $t=14$ s. - From 0 to 8 s: speed increases linearly from 0 to 9 m/s, area is a triangle. $$\text{Area}_1 = \frac{1}{2} \times 8 \times 9 = 36 \text{ m}$$ - From 8 to 14 s: speed constant at 9 m/s for 6 s. $$\text{Area}_2 = 9 \times 6 = 54 \text{ m}$$ - Total distance: $$36 + 54 = 90 \text{ m}$$ 4. **Part c: Average speed in first 7 seconds** - Speed increases linearly from 0 to $v$ at $t=7$ s. - Calculate speed at $t=7$ s using linear relation: $$v = 1.125 \times 7 = 7.875\text{ m/s}$$ - Average speed over 7 s is average of 0 and 7.875: $$\text{Average speed} = \frac{0 + 7.875}{2} = 3.9375\text{ m/s}$$ 5. **Part d: Solve $2^{x+3} = \frac{1}{16}$** - Rewrite $\frac{1}{16} = 2^{-4}$. - So: $$2^{x+3} = 2^{-4} \implies x + 3 = -4 \implies x = -7$$ 6. **Part e: Find $\int (6x^2 - 2x + 7) \, dx$** - Integrate each term separately: $$\int 6x^2 \, dx = 6 \times \frac{x^3}{3} = 2x^3$$ $$\int (-2x) \, dx = -2 \times \frac{x^2}{2} = -x^2$$ $$\int 7 \, dx = 7x$$ - Therefore: $$\int (6x^2 - 2x + 7) \, dx = 2x^3 - x^2 + 7x + C$$ \textbf{Final answers:} a) Acceleration = $1.125$ m/s$^2$ b) Distance = $90$ meters c) Average speed = $3.9375$ m/s d) $x = -7$ e) $2x^3 - x^2 + 7x + C$