1. **Problem statement:** A particle P moves in a straight line with velocity given by $$v = 0.5t \quad \text{for } 0 \leq t \leq 10,$$ $$v = 0.25t^{2} - 8t + 60 \quad \text{for } 10 \leq t \leq 20.$$ (a) Show there is an instantaneous change in acceleration at $t=10$. 2. **Formula and rules:** Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$: $$a = \frac{dv}{dt}.$$ To check for instantaneous change, find $a$ from both expressions at $t=10$ and compare. 3. **Calculate acceleration for $0 \leq t \leq 10$:** $$v = 0.5t \implies a = \frac{d}{dt}(0.5t) = 0.5.$$ At $t=10$, acceleration is $a = 0.5$. 4. **Calculate acceleration for $10 \leq t \leq 20$:** $$v = 0.25t^{2} - 8t + 60,$$ $$a = \frac{d}{dt}(0.25t^{2} - 8t + 60) = 0.5t - 8.$$ At $t=10$, $$a = 0.5 \times 10 - 8 = 5 - 8 = -3.$$ 5. **Compare accelerations at $t=10$:** From left side: $a=0.5$, from right side: $a=-3$. Since $0.5 \neq -3$, there is an instantaneous change in acceleration at $t=10$. --- 6. **(b) Find total distance covered in $0 \leq t \leq 20$:** Distance is the integral of the absolute value of velocity over time. 7. **Find position function $s(t)$ by integrating velocity:** For $0 \leq t \leq 10$, $$s(t) = \int 0.5t \, dt = 0.25t^{2} + C_1.$$ Assuming $s(0)=0$, then $C_1=0$, so $$s(t) = 0.25t^{2}.$$ At $t=10$, $$s(10) = 0.25 \times 10^{2} = 0.25 \times 100 = 25.$$ 8. **For $10 \leq t \leq 20$, integrate velocity:** $$s(t) = \int (0.25t^{2} - 8t + 60) dt = \frac{0.25}{3}t^{3} - 4t^{2} + 60t + C_2 = \frac{t^{3}}{12} - 4t^{2} + 60t + C_2.$$ 9. **Find $C_2$ using continuity of position at $t=10$:** $$s(10) = 25 = \frac{10^{3}}{12} - 4 \times 10^{2} + 60 \times 10 + C_2,$$ $$25 = \frac{1000}{12} - 400 + 600 + C_2,$$ $$25 = 83.3333 - 400 + 600 + C_2,$$ $$25 = 283.3333 + C_2,$$ $$C_2 = 25 - 283.3333 = -258.3333.$$ 10. **Position function for $10 \leq t \leq 20$ is:** $$s(t) = \frac{t^{3}}{12} - 4t^{2} + 60t - 258.3333.$$ 11. **Find position at $t=20$:** $$s(20) = \frac{8000}{12} - 4 \times 400 + 1200 - 258.3333 = 666.6667 - 1600 + 1200 - 258.3333 = 8.3334.$$ 12. **Check velocity sign to find if particle changes direction:** Velocity at $t=10$ is $$v(10) = 0.5 \times 10 = 5 > 0,$$ Velocity at $t=20$ is $$v(20) = 0.25 \times 400 - 8 \times 20 + 60 = 100 - 160 + 60 = 0.$$ Velocity is positive at $t=10$ and zero at $t=20$, check if velocity becomes negative in between. 13. **Find roots of $v = 0.25t^{2} - 8t + 60$ in $[10,20]$:** Solve $$0.25t^{2} - 8t + 60 = 0,$$ Multiply both sides by 4: $$t^{2} - 32t + 240 = 0,$$ Discriminant: $$\Delta = 32^{2} - 4 \times 1 \times 240 = 1024 - 960 = 64,$$ Roots: $$t = \frac{32 \pm 8}{2} = 20 \text{ or } 12.$$ 14. **Velocity zeros at $t=12$ and $t=20$ in $[10,20]$:** Velocity changes sign at $t=12$. 15. **Calculate position at $t=12$:** $$s(12) = \frac{12^{3}}{12} - 4 \times 12^{2} + 60 \times 12 - 258.3333 = 144 - 576 + 720 - 258.3333 = 29.6667.$$ 16. **Calculate distances between points:** From $t=0$ to $t=10$: distance = $s(10) - s(0) = 25 - 0 = 25$. From $t=10$ to $t=12$: distance = $s(12) - s(10) = 29.6667 - 25 = 4.6667$. From $t=12$ to $t=20$: velocity is negative (since it crosses zero at 12 and 20), so distance = $|s(20) - s(12)| = |8.3334 - 29.6667| = 21.3333$. 17. **Total distance covered:** $$25 + 4.6667 + 21.3333 = 51.$$ **Final answers:** (a) There is an instantaneous change in acceleration at $t=10$ because acceleration jumps from $0.5$ to $-3$. (b) The total distance covered by particle P in $0 \leq t \leq 20$ is approximately $51$ meters.