Acceleration at a given time

First find the velocity by differentiating $s(t) = 3t^5 - 5t^3$: $v(t) = s'(t) = 15t^4 - 15t^2$. Then differentiate again to find acceleration: $a(t) = v'(t) = 60t^3 - 30t$.

Substitute $t = 1$ into $a(t)$: $a(1) = 60(1)^3 - 30(1) = 60 - 30 = 30$. So the acceleration at $t=1$ is $30$ m/s².

We have $a(t) = 60t^3 - 30t = 30t(2t^2 - 1)$. Setting $a(t) = 0$ gives $t = 0$ or $t^2 = 1/2$, so $t = 0$ or $t = 1/\sqrt{2}$ (restricting to $t \geq 0$).

Testing signs on the intervals $t \geq 0$: For $0 < t < 1/\sqrt{2}$, pick $t = 0.5$: $a(0.5) = 30(0.5)(2(0.25)-1) = 30(0.5)(-0.5) < 0$, so $s(t)$ is concave down there. For $t > 1/\sqrt{2}$, pick $t=1$: $a(1) = 30 > 0$, so $s(t)$ is concave up. The second derivative changes sign at $t=1/\sqrt{2}$, so this is an inflection point. At $t=0$, we have $a(0) = 0$. Just to the right of $t=0$ (for small positive $t$), we have $a(t) < 0$ (concave down). Since we restrict to $t \geq 0$, the second derivative does not change sign when crossing $t=0$ within the domain $t \geq 0$, so $t=0$ is not an inflection point.

At $t = 1$, velocity is $v(1) = 15(1)^4 - 15(1)^2 = 15 - 15 = 0$, and acceleration is $a(1) = 30 > 0$. For $t$ slightly less than 1, $v(t) = 15t^4 - 15t^2 = 15t^2(t^2 - 1) < 0$ (since $t^2 < 1$), so the particle is moving in the negative direction. For $t$ slightly greater than 1, $v(t) = 15t^2(t^2 - 1) > 0$ (since $t^2 > 1$), so the particle is moving in the positive direction. Since the velocity is changing from negative to positive, and the magnitude is increasing due to positive acceleration, the particle is speeding up at $t = 1$.