# Question 3:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $J = \int_0^3 20te^{-t}dt$ | B1 | OR $\int_2^v 4dv = \int_0^3 20te^{-t}dt$ |
| $= \left[-20te^{-t}\right]_0^3 + \int_0^3 20e^{-t}dt$ | M1* | Use integration by parts (no limits required for M1 A1 A1) |
| Correctly applied to end of first stage | A1 | |
| $= \left[-20te^{-t} - 20e^{-t}\right]_0^3$ | A1 | |
| $= -3.98296\ldots + 20$ | M1 dep* | $20 - 80e^{-3}$ |
| $= 16.01703\ldots = 16.0$ Ns | A1 | Award only when identified as $J$ |
| $J = \Delta p$; $16.01703\ldots = 4v - 8$ | M1 | Working may be seen above |
| $v = 6.00425\ldots = 6.00$ ms$^{-1}$ | A1ft | $7 - 20e^{-3}$ |
| **[8]** | | |

**Alternative (find $v$ first):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\frac{dv}{dt} = 20te^{-t}$ | B1 | |
| $v = -5te^{-t} + \int 5e^{-t}dt$ | M1* A1 | M1 use integration by parts; A1 correctly applied to end of first stage |
| $v = -5te^{-t} - 5e^{-t}(+c)$ | A1 | |
| $t = 0, v = 2 \Rightarrow c = 7$ | M1 dep* | Use initial conditions |
| $v = 6.00425\ldots$ | A1 | |
| $J = 4(6.00425\ldots - 2) = 16.017\ldots$ | M1 A1ft | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = 20te^{-t} - \frac{1}{2}t = 0$ | B1 | |
| $t(20e^{-t} - \frac{1}{2}) = 0$, so $20e^{-t} - \frac{1}{2} = 0$ | | |
| $t = \ln 40$ | E1 | |
| $t = 0$ not applicable (since $t \geq 3$ so this is the only stationary point) | E1 | Stating $t = 0$ is not applicable; must clearly state $t \neq 0$ but no reason necessary; this mark can be awarded if explicitly stating $t \geq 3$ |
| $F$ and hence $a$ change sign (positive to negative) either side of $t = \ln 40$ | B1 | Any correct argument that $v$ is a maximum when $t = \ln 40$ |
| $4\frac{dv}{dt} = 20te^{-t} - \frac{1}{2}t$ | B1 | |
| $\int 4dv = \int\left(20te^{-t} - \frac{1}{2}t\right)dt$ | M1* | Separate and attempt to integrate; using $v$ from part (i) is M0 |
| $4v = -20e^{-t}(1+t) - \frac{1}{4}t^2(+c)$ | A2 | A1 for one error |
| When $t = 3$, $v = 6.00425\ldots$ | M1 dep* | Use initial conditions |
| $\Rightarrow c = 30\frac{1}{4}$ | | $c = 7\frac{9}{16}$ if $v = \ldots$ form used |
| When $t = \ln 40$, $v = 6.1259\ldots$ | M1 | Dependent on both previous M marks; substitute $t = \ln 40$ and solve for $v$ |
| $v = 6.13$ ms$^{-1}$ | A1 | cao |
| **[11]** | | |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = \sqrt{121 - 80e^{-T}(1+T)}$ | B1 | $80e^{-11}(1+11)$ as being small (oe) |
| Justify $T$ is approximately $11$ | E1 | Considers $e^{-T}$ becoming small |
| **[2]** | | |

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. starting value of $11$: $10.99927117\ldots$; $10.99927069\ldots$ | M1 | A suitable method |
| so $t = 10.9993$ to 4 dp | M1 | Correct working seen; must obtain answer to at least 4dp |
| | A1 | Justifiable conclusion from their working ($10.9993$ only; must be 4dp) |
| **[3]** | | |

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