Question Statement

A car is moving along a straight road. When $t=0\text{s}$, the car passes a point $A$ with velocity $10\text{m/s}$ and this velocity is maintained until $t=30\text{s}$. The driver then applies the brakes and the car decelerates uniformly, coming to rest at point $B$ when $t=42\text{s}$.

(i) Sketch a velocity-time graph to illustrate the motion of the car.

(ii) Find the distance from $A$ to $B$.

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Background and Explanation

This problem involves analyzing motion with constant velocity followed by uniform deceleration. You will need to apply the equations of motion for uniformly accelerated objects and understand that the area under a velocity-time graph represents displacement.

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Solution

Part (i): Velocity-Time Graph

The motion consists of two distinct phases:

• Phase 1 (0 to 30 s): Constant velocity of $10\text{m/s}$, represented by a horizontal line.
• Phase 2 (30 to 42 s): Uniform deceleration from $10\text{m/s}$ to rest ($0\text{m/s}$), represented by a straight line with negative slope.

The sketch below illustrates this motion, showing velocity on the vertical axis and time on the horizontal axis:

Part (ii): Distance from A to B

To find the total distance, we calculate the distance traveled during each phase separately and sum them.

Phase 1: Constant Velocity ($t=0$ to $t=30\text{s}$)

During this phase, the car moves at constant speed: $v=10\text{m/s},t=30\text{s}$

Using the formula for distance at constant velocity: $S_1=v\times t=10\times 30=300\text{m}$

Phase 2: Uniform Deceleration ($t=30\text{s}$ to $t=42\text{s}$)

For the braking phase:

• Initial velocity: $v_i=10\text{m/s}$ (velocity at $t=30\text{s}$)
• Final velocity: $v_f=0\text{m/s}$ (comes to rest at point $B$)
• Time duration: $t=42-30=12\text{s}$

First, calculate the acceleration using $v_f=v_i+at$: $0=10+a(12)$ $a=\frac{-10}{12}=-\frac{5}{6}{\text{m/s}}^2$

Now calculate the distance traveled during braking using $s=v_{i}t+\frac{1}{2}a{t}^2$: $S_2=10\times 12+\frac{1}{2}\left(-\frac{5}{6}\right)(12{)}^2$ $S_2=120+\frac{1}{2}\left(-\frac{5}{6}\right)(144)$ $S_2=120-60=60\text{m}$

Total Distance: $\text{Total distance}=S_1+S_2=300+60=360\text{m}$

Therefore, the distance from $A$ to $B$ is $\boxed{360\text{m}}$.

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Key Formulas or Methods Used

• Distance at constant velocity: $s=v\times t$
• First equation of motion: $v=u+at$ (or $v_f=v_i+at$)
• Second equation of motion: $s=ut+\frac{1}{2}a{t}^2$ (or $s=v_{i}t+\frac{1}{2}a{t}^2$)
• Total distance = sum of distances from individual motion phases

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Summary of Steps

1. Divide the motion into two phases: constant velocity ($0$–$30\text{s}$) and uniform deceleration ($30$–$42\text{s}$)
2. Calculate distance $S_1$ for the first phase using $s=vt$
3. Determine the time interval for the deceleration phase: $t=42-30=12\text{s}$
4. Calculate acceleration using $v_f=v_i+at$ with final velocity $v_f=0$
5. Calculate distance $S_2$ for the second phase using $s=v_{i}t+\frac{1}{2}a{t}^2$
6. Add $S_1$ and $S_2$ to find the total distance from $A$ to $B$: $300\text{m}+60\text{m}=360\text{m}$