Standard +0.3 This is a straightforward M1 kinematics question requiring students to: (1) differentiate velocity to find acceleration, (2) solve a linear equation to find when a=0, (3) integrate velocity to find displacement, and (4) add the initial position. While it involves multiple steps and careful attention to the initial condition, each step uses standard techniques with no conceptual challenges or novel problem-solving required. Slightly above average difficulty due to the multi-step nature and need to track the initial position correctly.
\includegraphics{figure_5}
A toy car is moving along the straight line \(Ox\), where O is the origin. The time \(t\) is in seconds. At time \(t = 0\) the car is at A, 3 m from O as shown in Fig. 5. The velocity of the car, \(v\) m s\(^{-1}\), is given by
$$v = 2 + 12t - 3t^2.$$
Calculate the distance of the car from O when its acceleration is zero. [8]
Differentiation, at least one term correct. Follow their \(a\)
\(a = 0\) gives \(t = 2\)
\(x = \int(2t + 12t - 3t^2)dx\)
M1
Integration indefinite or definite, at least one term correct. Correct. Need not be simplified. Allow as definite integral. Ignore \(C\) or limits.
\(2t + 6t^2 - t^3 + C\)
A1
Allow \(x = \pm 3\) or argue it is \(\int_0\) from A then \(\pm 3\)
\(x = 3\) when \(t = 0\)
M1
so \(3 = C\) and
Award if seen WWW or \(x = 2t + 6t^2 - t^3\) seen with \(+3\) added later. FT their \(t\) and their \(x\) if obtained by integration but not if \(-3\) obtained instead of \(+3\). [If 20 m seen WWW for displacement award SC6] [Award SC1 for position if constant acceleration used for displacement and then \(+3\) applied]
\(x = 2t + 6t^2 - t^3 + 3\)
A1
\(x(4) = 4 + 24 - 8 + 3 = 23\) m
B1
Sub: 8
$a = 12 - 6t$ | M1, A1, F1 | Differentiation, at least one term correct. Follow their $a$
$a = 0$ gives $t = 2$ | |
$x = \int(2t + 12t - 3t^2)dx$ | M1 | Integration indefinite or definite, at least one term correct. Correct. Need not be simplified. Allow as definite integral. Ignore $C$ or limits.
$2t + 6t^2 - t^3 + C$ | A1 | Allow $x = \pm 3$ or argue it is $\int_0$ from A then $\pm 3$
$x = 3$ when $t = 0$ | M1 |
so $3 = C$ and | | Award if seen WWW or $x = 2t + 6t^2 - t^3$ seen with $+3$ added later. FT their $t$ and their $x$ if obtained by integration but not if $-3$ obtained instead of $+3$. [If 20 m seen WWW for displacement award SC6] [Award SC1 for position if constant acceleration used for displacement and then $+3$ applied]
$x = 2t + 6t^2 - t^3 + 3$ | A1 |
$x(4) = 4 + 24 - 8 + 3 = 23$ m | B1 |
| Sub: 8
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\includegraphics{figure_5}
A toy car is moving along the straight line $Ox$, where O is the origin. The time $t$ is in seconds. At time $t = 0$ the car is at A, 3 m from O as shown in Fig. 5. The velocity of the car, $v$ m s$^{-1}$, is given by
$$v = 2 + 12t - 3t^2.$$
Calculate the distance of the car from O when its acceleration is zero. [8]
\hfill \mbox{\textit{OCR MEI M1 2008 Q5 [8]}}