Question Statement

A car moving on a straight road is modelled as a particle moving along the $x$-axis, and its acceleration $am/{\mathrm{s}}^2$, after a given instant $t$, is given by:

$a=\{\begin{array}{ll}4-\frac{1}{2}t& 0\le t\le 8\\ 0& t>8\end{array}$

The car starts from rest.

(i) Find a similar expression for the velocity of the car, as that of its acceleration.

(ii) Find the time it takes for the car to reach its maximum speed.

(iii) Show that the displacement of the car from the origin is given by:

$S=\{\begin{array}{ll}2t^2-\frac{1}{12}{t}^3& 0\le t\le 8\\ 16t-\frac{128}{3}& t>8\end{array}$

(iv) Calculate the time it takes the car to cover the first $1000\mathrm{m}$.

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Background and Explanation

This problem involves kinematics with variable acceleration, requiring integration of the acceleration function to obtain velocity, and integration of velocity to obtain displacement. Since the acceleration is defined piecewise, you must handle each time interval separately and ensure continuity of velocity and displacement at the boundary $t=8$.

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Solution

Part (i): Finding the Velocity Expression

To find velocity, we integrate the acceleration with respect to time. Since the car starts from rest, we use the initial condition $v=0$ when $t=0$.

For $0\le t\le 8$:

$\begin{array}{rl}v& =\int adt\\ & =\int \left(4-\frac{1}{2}t\right)dt\\ & =4t-\frac{1}{4}{t}^2+C\end{array}$

Applying the initial condition: when $t=0$, $v=0$:

$0=0-0+C\Rightarrow C=0$

Thus, for $0\le t\le 8$: $v=4t-\frac{1}{4}{t}^2$

For $t>8$: Since $a=0$, the velocity remains constant at the value attained at $t=8$:

$v=4(8)-\frac{1}{4}(8{)}^2=32-\frac{64}{4}=32-16=16m/s$

Therefore, the velocity is:

$v=\{\begin{array}{ll}4t-\frac{1}{4}{t}^2& 0\le t\le 8\\ 16& t>8\end{array}$

Part (ii): Time to Reach Maximum Speed

The car reaches maximum speed when acceleration drops to zero (transitioning from speeding up to constant velocity).

Setting $a=0$ in the first interval:

$\begin{array}{rl}4-\frac{1}{2}t& =0\\ \frac{1}{2}t& =4\\ t& =8\mathrm{s}\end{array}$

At $t=8\mathrm{s}$, the velocity reaches its maximum value of $16m/s$ and remains constant thereafter.

Part (iii): Deriving the Displacement Expression

Displacement is found by integrating the velocity function.

For $0\le t\le 8$:

$\begin{array}{rl}s& =\int vdt\\ & =\int \left(4t-\frac{1}{4}{t}^2\right)dt\\ & =\frac{4t^2}{2}-\frac{1}{4}\cdot \frac{t^3}{3}+c\\ & =2t^2-\frac{1}{12}{t}^3+c\end{array}$

Using initial condition $s=0$ when $t=0$: $c=0$.

Thus: $s=2t^2-\frac{1}{12}{t}^3$

For $t>8$: We calculate the displacement by adding the distance traveled during constant velocity motion to the displacement at $t=8$.

First, find displacement at $t=8$: $s(8)=2(8{)}^2-\frac{1}{12}(8{)}^3=128-\frac{512}{12}=128-\frac{128}{3}=\frac{256}{3}\mathrm{m}$

For $t>8$, using definite integration from $8$ to $t$: $\begin{array}{rl}s& ={\int }_8^{t}16d\tau +s(8)\\ & =16(t-8)+\frac{256}{3}\\ & =16t-128+\frac{256}{3}\\ & =16t-\frac{384}{3}+\frac{256}{3}\\ & =16t-\frac{128}{3}\end{array}$

Therefore, the displacement is:

$S=\{\begin{array}{ll}2t^2-\frac{1}{12}{t}^3& 0\le t\le 8\\ 16t-\frac{128}{3}& t>8\end{array}$

Part (iv): Time to Cover 1000 m

First, check if the car reaches $1000\mathrm{m}$ during the acceleration phase ($t\le 8$): $s(8)=\frac{256}{3}\approx 85.33\mathrm{m}$

Since $85.33\mathrm{m}<1000\mathrm{m}$, the car reaches $1000\mathrm{m}$ during the constant velocity phase ($t>8$).

Using the second displacement equation: $\begin{array}{rl}16t-\frac{128}{3}& =1000\\ 16t& =1000+\frac{128}{3}\\ 16t& =\frac{3000+128}{3}=\frac{3128}{3}\\ t& =\frac{3128}{3\times 16}=\frac{3128}{48}=\frac{391}{6}\mathrm{s}\\ t& \approx 65.17\mathrm{s}\end{array}$

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Key Formulas or Methods Used

• $v=\int adt$ (Integration of acceleration yields velocity)
• $s=\int vdt$ (Integration of velocity yields displacement)
• Constant of integration determined by initial/boundary conditions
• Continuity conditions at $t=8$ to connect piecewise functions
• Maximum velocity occurs when $a=0$ (for motion with decreasing acceleration)

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Summary of Steps

1. Find Velocity: Integrate $a=4-\frac{1}{2}t$ to get $v=4t-\frac{1}{4}{t}^2$ for $0\le t\le 8$, then use continuity to establish $v=16$ for $t>8$.
2. Find Maximum Speed Time: Set $a=0$ and solve $4-\frac{1}{2}t=0$ to get $t=8\mathrm{s}$.
3. Find Displacement: Integrate velocity piecewise; for $t>8$, add the constant velocity contribution to the displacement at $t=8$ ($\frac{256}{3}\mathrm{m}$).
4. Calculate Time for 1000 m: Verify that $1000\mathrm{m}$ occurs after $t=8$, then solve $16t-\frac{128}{3}=1000$ to obtain $t=\frac{391}{6}\mathrm{s}\approx 65.17\mathrm{s}$.