The displacement (distance), s, in metres, travelled by car A over time (t) in seconds, after the brakes were applied until the car stopped, is represented by a formula s = 30t - 3t² 8.1 If $ \frac{ds}{dt} = 0 $, determine the time t (in seconds). HINT: $ \frac{ds}{dt} $ is the velocity of the car in m/s. 8.2 Determine: 8.2.1 The velocity, in kilometres per hour, at the time when the brakes were first applied 8.2.2 The maximum distance travelled by car A before it stopped 8.3 State, with reasons, whether car A will collide with stationary car B, which is 70 m directly in front of car A, after the brakes have been applied.

NSC Technical Mathematics Differential Calculus and Integration: The displacement (distance), s,

The displacement (distance), s, in metres, travelled by car A over time (t) in seconds, after the brakes were applied until the car stopped, is represented by a formula s = 30t - 3t² 8.1 If $ \frac{ds}{dt} = 0 $, determine the time t (in seconds) - NSC Technical Mathematics - Question 8 - 2021 - Paper 1

Question 8

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The displacement (distance), s, in metres, travelled by car A over time (t) in seconds, after the brakes were applied until the car stopped, is represented by a form... show full transcript

Worked Solution & Example Answer:The displacement (distance), s, in metres, travelled by car A over time (t) in seconds, after the brakes were applied until the car stopped, is represented by a formula s = 30t - 3t² 8.1 If $ \frac{ds}{dt} = 0 $, determine the time t (in seconds) - NSC Technical Mathematics - Question 8 - 2021 - Paper 1

Step 1

If $ \frac{ds}{dt} = 0 $, determine the time t (in seconds).

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Answer

To find the time when the car stops, we set the velocity to zero:

$\frac{ds}{dt} = 30 - 6t$

Setting this equal to zero gives:

$30 - 6t = 0$

Solving for t:

$6t = 30$ $t = 5 \text{ seconds}$

Step 2

The velocity, in kilometres per hour, at the time when the brakes were first applied.

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Answer

When the brakes were first applied, we need to evaluate ( \frac{ds}{dt} ) at ( t = 0 ):

$\frac{ds}{dt} = 30 - 6(0) = 30 \text{ m/s}$

To convert this to kilometres per hour:

$30 \text{ m/s} \times \frac{3600 \text{ s}}{1000 \text{ m}} = 108 \text{ km/h}$

Step 3

The maximum distance travelled by car A before it stopped.

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Answer

To find the maximum distance, we substitute ( t = 5 ) seconds into the displacement formula:

$s = 30(5) - 3(5)^2$

Calculating this gives:

$s = 150 - 75 = 75 \text{ metres}$

Step 4

State, with reasons, whether car A will collide with stationary car B, which is 70 m directly in front of car A, after the brakes have been applied.

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Answer

Since car A travels a maximum distance of 75 m after the brakes are applied, and car B is located 70 m in front, the two cars will indeed collide. Because 75 m > 70 m, car A covers more distance than the gap to car B.

If \( \frac{ds}{dt} = 0 \), determine the time t (in seconds).

The velocity, in kilometres per hour, at the time when the brakes were first applied.

The maximum distance travelled by car A before it stopped.

State, with reasons, whether car A will collide with stationary car B, which is 70 m directly in front of car A, after the brakes have been applied.

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