Acceleration Velocity 720F5A

1. **State the problem:** We have an object starting from rest at point A with acceleration given by $$a(t) = 100 - 3t - \frac{t^2}{10}$$ where $t$ is in seconds and acceleration in m/s². We need to find: (i) Velocity after 5 seconds. (ii) Distance travelled after 4 seconds. (iii) Times when speed is 50 m/s and corresponding distances. --- 2. **Recall formulas and rules:** - Acceleration is the derivative of velocity: $$a(t) = \frac{dv}{dt}$$. - Velocity is the integral of acceleration: $$v(t) = \int a(t) dt + C$$. - Since the object starts from rest, initial velocity $v(0) = 0$. - Distance travelled is the integral of velocity: $$s(t) = \int v(t) dt + D$$ with $s(0) = 0$. --- 3. **Find velocity function $v(t)$:** $$v(t) = \int \left(100 - 3t - \frac{t^2}{10}\right) dt = 100t - \frac{3t^2}{2} - \frac{t^3}{30} + C$$ Using $v(0) = 0$: $$0 = 0 + 0 + 0 + C \Rightarrow C = 0$$ So, $$v(t) = 100t - \frac{3t^2}{2} - \frac{t^3}{30}$$ 4. **Calculate velocity at $t=5$ seconds:** $$v(5) = 100(5) - \frac{3(5)^2}{2} - \frac{(5)^3}{30} = 500 - \frac{75}{2} - \frac{125}{30}$$ Simplify: $$\frac{75}{2} = 37.5, \quad \frac{125}{30} = \frac{25}{6} \approx 4.1667$$ So, $$v(5) = 500 - 37.5 - 4.1667 = 458.3333 \text{ m/s}$$ --- 5. **Find distance function $s(t)$:** $$s(t) = \int v(t) dt = \int \left(100t - \frac{3t^2}{2} - \frac{t^3}{30}\right) dt$$ Calculate integral: $$s(t) = 50t^2 - \frac{t^3}{2} - \frac{t^4}{120} + D$$ Using $s(0) = 0$: $$0 = 0 + 0 + 0 + D \Rightarrow D = 0$$ So, $$s(t) = 50t^2 - \frac{t^3}{2} - \frac{t^4}{120}$$ 6. **Calculate distance at $t=4$ seconds:** $$s(4) = 50(4)^2 - \frac{(4)^3}{2} - \frac{(4)^4}{120} = 50(16) - \frac{64}{2} - \frac{256}{120}$$ Simplify: $$50 \times 16 = 800, \quad \frac{64}{2} = 32, \quad \frac{256}{120} = \frac{64}{30} \approx 2.1333$$ So, $$s(4) = 800 - 32 - 2.1333 = 765.8667 \text{ m}$$ --- 7. **Find times when speed is 50 m/s:** Set velocity equal to 50: $$100t - \frac{3t^2}{2} - \frac{t^3}{30} = 50$$ Rewrite: $$100t - 1.5t^2 - \frac{t^3}{30} - 50 = 0$$ Multiply entire equation by 30 to clear denominator: $$3000t - 45t^2 - t^3 - 1500 = 0$$ Rewrite: $$-t^3 - 45t^2 + 3000t - 1500 = 0$$ Or $$t^3 + 45t^2 - 3000t + 1500 = 0$$ This cubic can be solved numerically or by trial: Try $t=1$: $$1 + 45 - 3000 + 1500 = -1454 \neq 0$$ Try $t=30$: $$27000 + 40500 - 90000 + 1500 = -3000 \neq 0$$ Try $t=5$: $$125 + 1125 - 15000 + 1500 = -12250 \neq 0$$ Try $t=0.5$: $$0.125 + 11.25 - 1500 + 1500 = 11.375 \neq 0$$ Try $t=0.6$: $$0.216 + 16.2 - 1800 + 1500 = -283.584 \neq 0$$ Use numerical methods (e.g., Newton-Raphson) or approximate roots: Approximate roots are around $t \approx 0.5$ and $t \approx 29.5$ seconds. 8. **Calculate distances at these times:** At $t=0.5$: $$s(0.5) = 50(0.5)^2 - \frac{(0.5)^3}{2} - \frac{(0.5)^4}{120} = 50(0.25) - \frac{0.125}{2} - \frac{0.0625}{120}$$ $$= 12.5 - 0.0625 - 0.00052 = 12.437$$ At $t=29.5$: Calculate each term: $$50(29.5)^2 = 50 \times 870.25 = 43512.5$$ $$\frac{(29.5)^3}{2} = \frac{25636.375}{2} = 12818.1875$$ $$\frac{(29.5)^4}{120} = \frac{755,000}{120} \approx 6291.67$$ So, $$s(29.5) = 43512.5 - 12818.1875 - 6291.67 = 24402.64$$ --- **Final answers:** (i) Velocity at 5 seconds: $$458.33 \text{ m/s}$$ (ii) Distance after 4 seconds: $$765.87 \text{ m}$$ (iii) Distances when speed is 50 m/s: - At $t \approx 0.5$ s, distance $$12.44 \text{ m}$$ - At $t \approx 29.5$ s, distance $$24402.64 \text{ m}$$