Question

Point A is launched horizontally away from a wall and a ( t ) a(t) represents Point A's distance from the wall (in meters) in terms of the number of seconds t t since the point was launched. Point D is launched from a platform that places it 1 meters from the wall, and Point D travels 1.5 times as fast as Point A. AD Click here to launch the points. Suppose d ( t ) d(t) represents Point D's distance from the wall (in meters) in terms of the number of seconds t t since the points were launched. Write a function formula for d d using the function a a.

Answer 1

Answer:

d(t) = 1 +1.5*a(t)

Step-by-step explanation:

Since point D is launched from a platform that places it 1 meters from the wall, we have  

d(0) = 1

Since point A is launched directly from the wall,  

a(0) = 0

but besides, point D travels 1.5 times as fast as point A, so

d(t) = 1 +1.5*a(t)

and this the function required.

Answer 2

The function formula for 'd' using the function 'a' for considered situation is $d(t) = 1 + 1.5 \times a(t)$

How to form mathematical expression from the given description?

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.

How to find the speed of an object?

If the object is going linearly, and at constant speed, then the speed of that object is given by the distance it traveled to the time it took to travel that distance.

If the object traveled D distance in T units time, then that object's speed is

$Speed = S = \dfrac{\: Distance \: traveled}{\: Time \: taken} = \dfrac{D}{T} \: \rm unit \: length/unit \: time$

Now, for the given case, the points are moving horizontally, thus, on a linear path.

For points A and D, their distance functions are a(t) and d(t) respectively.

Supposing both points' speeds are constant for any time, for t = 0, as we are given that: a(0) = 0(distance from wall is 0).

and d(0) = 1 meter(it is 1 meter away from wall), so initial distance = 0

After time t, we get their positions as a(t) and d(t) respectively.

Therefore, their speeds are calculated as:

For the time interval from 0 seconds to t seconds, we get:

• Speed of A = $S_A = \dfrac{a(t) - a(0)}{t-0} = \dfrac{a(t)}{t} \: \rm m/s$

Similarly, we get:

• Speed of D = $S_D = \dfrac{d(t) - d(0)}{t-0} = \dfrac{d(t) -1}{t} \: \rm m/s$

it is because the distance traveled is the difference of the distance function for final and initial time. And the time taken is the difference between final and initial time.

It is specified that D is traveling 1.5 times faster than A. That means,.

$S_D = 1.5 \times S_A$

or,

$\dfrac{d(t) - 1}{t} = 1.5 \times \dfrac{a(t)}{t}\\\\d(t) = 1 + 1.5 \times a(t)$

Therefore, the function formula for 'd' using the function 'a' for considered situation is $d(t) = 1 + 1.5 \times a(t)$

Learn more about speed here:

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