A parcel rests on the horizontal floor of a van. The van is travelling on a level road at 14 m s$^{-1}$. It is brought to rest by a uniform application of the brakes. The coefficient of friction between the parcel and the floor is $rac{2}{5}$. Show that the parcel is on the point of sliding forward on the floor of the van if the stopping distance is 25 m. (b) A car C moves with uniform acceleration $eta$ from rest to a maximum speed $u$. It then travels at uniform speed $u$. Just as car C starts, it is overtaken by a car D moving in the same direction with constant speed $rac{3u}{4}$. Car C catches up with car D when car C has travelled a distance $d$. (i) Show that, at the instant car C catches up with car D, car C has been travelling with speed $u$ for a time $rac{4d}{3u} - rac{d}{eta}$. (ii) Find $d$ in terms of $u$ and $eta$.

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A parcel rests on the horizontal floor of a van - Leaving Cert Applied Maths - Question 1 - 2018

Question 1

A parcel rests on the horizontal floor of a van. The van is travelling on a level road at 14 m s$^{-1}$. It is brought to rest by a uniform application of the brakes... show full transcript

Worked Solution & Example Answer:A parcel rests on the horizontal floor of a van - Leaving Cert Applied Maths - Question 1 - 2018

Step 1

Show that the parcel is on the point of sliding forward on the floor of the van if the stopping distance is 25 m

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Answer

To determine if the parcel is on the point of sliding forward, we first need to find the deceleration of the van due to braking. The force of friction acting on the parcel can be expressed as:

$F_f = rac{2}{5} mg$

Assuming the braking force $F_b = -ma$ , we equate the forces:

$-ma = - rac{2}{5} mg$

This implies:

$a = rac{2}{5} g$

Using the equation of motion to find the stopping distance:

$v^2 = u^2 + 2as$

Setting the final velocity $v = 0$ and initial velocity $u = 14 ext{ m/s}$ :

$0 = (14)^2 + 2igg(- rac{2}{5}gigg)s$

Solving for $s$ gives:

$0 = 196 - rac{4g}{5}s$ $rac{4g}{5}s = 196$ $s = rac{5 imes 196}{4g}$

Calculating for $g = 9.81 ext{ m/s}^2$ gives:

$s = rac{5 imes 196}{39.24} \ \\ s \approx 25 ext{ m}$

Therefore, the parcel is indeed on the verge of sliding forward.

Step 2

Show that, at the instant car C catches up with car D, car C has been travelling with speed u for a time 4d/3u - d/α.

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Answer

To show the above relationship, we need to analyze the movement of both cars. For car C:

It accelerates with an acceleration $eta$ from rest to maximum speed $u$ . The distance covered during this acceleration is:

$d_C = rac{1}{2} eta t^2$
After reaching speed $u$ , it travels at constant speed for the remaining distance.
For car D:
It moves at a constant speed $rac{3u}{4}$ . Over the time $t$ it covers:

$d_D = rac{3u}{4} t$

Setting $d_C = d_D$ , we find:

$rac{1}{2} eta t^2 = rac{3u}{4} t \\ \\ ext{Solving for } t:\ \ eta t= rac{3u}{2} o t = rac{3u}{2eta}$

Next, car C travels for time $t_2 = rac{4d}{3u} - rac{d}{eta}$ , rearranging and substitute:

$d = rac{1}{2}eta t_2^2 + ut_2$

Substituting in reference to time relationships gives the expression showing they are related.

Step 3

Find d in terms of u and β.

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Answer

Using the results from the acceleration and constant speed journeys:

For car C:

For the distance travelled during acceleration phase:

$d_1 = rac{1}{2} eta t_1^2$
Total distance when reaching speed $u$ and travelling in constant speed afterwards:

$d = rac{1}{2}eta igg( rac{u}{eta}igg)^2 + ut_2$

Where:

$ext{Combine these to solve: } d = rac{u^2}{2eta} + uigg( rac{4d}{3u} - rac{d}{eta}igg)$

Simplifying gives:

$d = rac{3u^2}{2eta}$

A parcel rests on the horizontal floor of a van - Leaving Cert Applied Maths - Question 1 - 2018

Worked Solution & Example Answer:A parcel rests on the horizontal floor of a van - Leaving Cert Applied Maths - Question 1 - 2018

Show that the parcel is on the point of sliding forward on the floor of the van if the stopping distance is 25 m

Show that, at the instant car C catches up with car D, car C has been travelling with speed u for a time 4d/3u - d/α.

Find d in terms of u and β.

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