A particle $P$ is moving along the $x$-axis. At time $t$ seconds ($t \geq 0$), the velocity of $P$ is
$ v \, \text{m/s}, \quad \text{where } v = t^2 - 10t + 28. $
(a) Find the velocity of $P$ when $t = 1$
Substitute $t = 1$ into $v = t^2 - 10t + 28$:
$ v = (1)^2 - 10(1) + 28 = 1 - 10 + 28 = 19. $
The velocity of $P$ when $t = 1$ is $19 \, \text{m/s}$.
(b) Find the distance of $P$ from the origin when $t = 5$
The position of $P$, $x(t)$, is obtained by integrating $v = t^2 - 10t + 28$:
$ x(t) = \int (t^2 - 10t + 28) \, dt = \frac{t^3}{3} - 5t^2 + 28t + C. $
When $t = 3$, the distance of $P$ from the origin is $x(3) = 24$:
$ x(3) = \frac{(3)^3}{3} - 5(3)^2 + 28(3) + C = 24. $
Simplify:
$ x(3) = \frac{27}{3} - 45 + 84 + C = 24. $
$ 9 - 45 + 84 + C = 24. $
$ C = 24 - 48 = -24. $
Thus, the position function is:
$ x(t) = \frac{t^3}{3} - 5t^2 + 28t - 24. $
Substitute $t = 5$ into $x(t)$ to find the distance from the origin:
$ x(5) = \frac{(5)^3}{3} - 5(5)^2 + 28(5) - 24. $
$ x(5) = \frac{125}{3} - 125 + 140 - 24. $
$ x(5) = \frac{125}{3} - 9. $
$ x(5) = \frac{125 - 27}{3} = \frac{98}{3} \approx 32.67. $
The distance of $P$ from the origin when $t = 5$ is approximately $32.67 \, \text{m}$.
(c) Find the acceleration of $P$ when $t = 9$
The acceleration of $P$ is the derivative of $v$:
$ a(t) = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 10t + 28) = 2t - 10. $
Substitute $t = 9$:
$ a(9) = 2(9) - 10 = 18 - 10 = 8. $
The acceleration of $P$ when $t = 9$ is $8 \, \text{m/s}^2$.
(d) (i) Show that there are no values of $t$ for which $P$ is instantaneously at rest
A particle is at rest when $v = 0$. Solve:
$ t^2 - 10t + 28 = 0. $
The discriminant of this quadratic equation is:
$ \Delta = (-10)^2 - 4(1)(28) = 100 - 112 = -12. $
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, there are no values of $t$ for which $P$ is instantaneously at rest.
(d) (ii) Find the least magnitude of the velocity of $P$
The velocity function is $v = t^2 - 10t + 28$. To find the least magnitude, minimize $|v|$. The critical points occur where $\frac{dv}{dt} = 0$:
$ \frac{dv}{dt} = 2t - 10 = 0 \implies t = 5. $
Evaluate $v(t)$ at $t = 5$:
$ v(5) = (5)^2 - 10(5) + 28 = 25 - 50 + 28 = 3. $
The velocity is always positive (since $v > 0$ for all $t \geq 0$), so the least magnitude of the velocity is:
$ \boxed{3 \, \text{m/s}}. $
Final Answers:
- (a) $v = 19 \, \text{m/s}$.
- (b) Distance from origin when $t = 5$: $\approx 32.67 \, \text{m}$.
- (c) Acceleration when $t = 9$: $8 \, \text{m/s}^2$.
- (d) (i) No values of $t$ for which $P$ is instantaneously at rest.
- (d) (ii) Least magnitude of velocity: $3 \, \text{m/s}$.
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