# Thread: speed and dist qn

1. ## speed and dist qn

can anyone explain to me wat i am doing?

for qn part (b), i got it correct when i did it last year but now when i look back, i dont understand what i am doing.

this is wat i did:

dist travelled by bicycle:2d
dist travelled by bicycle:d

T X 5 = 2d
0.5 X T X T/5 = d (i dont understand this step)
T^2/10 = d
T^2 / 5 = 5T
T=25.

The above method is correct as my teacher marked it last year already. Now i just need someone to remind me of why i did this. Thanks

2. Originally Posted by helloying
can anyone explain to me wat i am doing?

for qn part (b), i got it correct when i did it last year but now when i look back, i dont understand what i am doing.

this is wat i did:

dist travelled by bicycle:2d
dist travelled by bicycle:d

T X 5 = 2d
0.5 X T X T/5 = d (i dont understand this step)
T^2/10 = d
T^2 / 5 = 5T
T=25.

The above method is correct as my teacher marked it last year already. Now i just need someone to remind me of why i did this. Thanks
the step you do not understand is the standard formula for displacement for an object starting from rest ...

$d = \frac{1}{2}at^2$

3. Originally Posted by skeeter
the step you do not understand is the standard formula for displacement for an object starting from rest ...

$d = \frac{1}{2}at^2$
oh ok. But this is a bit out the the syllabus.what is in the tb is that the dist travelled is the area underneath the speed-time graph.i can only use this knowledge to solve the qn. i know there are many things the tb didnt teach.Can u kindly explain to me how to solve the qn using the simple knowledge?

4. for the first 40 seconds, the bus' velocity is $v(t) = at = \frac{1}{5}t$

area of a triangle, $A = \frac{1}{2} bh$ , is the displacement for the bus ...

$d = \frac{1}{2}t \cdot v(t)$

$d = \frac{1}{2}t \cdot \frac{1}{5}t$

5. In more advanced mathematics (Calculus) we would say that the total distance traveled is the integral of the velocity function. That is also the "area under the curve" (you say you know that) and here, that's easy- it is the area of a trapezoid with bases of length 100 and 40 and height 8: $\frac{100+ 40}{2}(8)= 70(8)= 560$ m. Perhaps your tm thought you had learned the area for a trapezoid long ago. Or you can do what skeeter suggested in his last post: the trapezoid can be seen as two triangles and a rectangle. Find the area of each and add.

The bus has three different accelerations, one of which is 0. Are you asked for all three or the average acceleration. If "average" this is trivial: it starts with 0 velocity and ends with 0 velocity!

Finally, for the last part, you are told that T< 40 so you are only looking at the first part of the graph. The bicycle travels at a constant 5 mph (another of those damn bicyclists who doesn't stop for red lights!) so after time T has gone 5t miles. That's exactly the same as saying that if you drop a perpendicular at time T, you have a rectangle of height 5 and base T and so area 5T. The bus, during that time gets up to a speed of 8 m/s in 40 s and so has an acceleration of 40/8= 5m/s^2. The distance gone by the bus is a triangle with base T and height 5T. Find the area of that triangle, set it equal to 5T and solve for T.

6. Originally Posted by HallsofIvy
In more advanced mathematics (Calculus) we would say that the total distance traveled is the integral of the velocity function. That is also the "area under the curve" (you say you know that) and here, that's easy- it is the area of a trapezoid with bases of length 100 and 40 and height 8: $\frac{100+ 40}{2}(8)= 70(8)= 560$ m. Perhaps your tm thought you had learned the area for a trapezoid long ago. Or you can do what skeeter suggested in his last post: the trapezoid can be seen as two triangles and a rectangle. Find the area of each and add.

The bus has three different accelerations, one of which is 0. Are you asked for all three or the average acceleration. If "average" this is trivial: it starts with 0 velocity and ends with 0 velocity!

Finally, for the last part, you are told that T< 40 so you are only looking at the first part of the graph. The bicycle travels at a constant 5 mph (another of those damn bicyclists who doesn't stop for red lights!) so after time T has gone 5t miles. That's exactly the same as saying that if you drop a perpendicular at time T, you have a rectangle of height 5 and base T and so area 5T. The bus, during that time gets up to a speed of 8 m/s in 40 s and so has an acceleration of 40/8= 5m/s^2. The distance gone by the bus is a triangle with base T and height 5T. Find the area of that triangle, set it equal to 5T and solve for T.
Sorry i still dont quite get the last part. Why is the height of the triangle 5T?Is it because of its accelearation? what is the relationship between them? And why should i set it equal to 5T? isnt it suppose to be half 5T because the qn say it is twice the dist.

7. Originally Posted by helloying
Sorry i still dont quite get the last part. Why is the height of the triangle 5T?Is it because of its accelearation? what is the relationship between them? And why should i set it equal to 5T? isnt it suppose to be half 5T because the qn say it is twice the dist.
Actually it's NOT. I accidently reversed a fraction.

Do this: at t= T (between 0 and 40), draw a vertical line up to the graph and draw a vertical line from t= 40 up to that line. You now have two right triangles with exactly the same angles- and thus are similar triangles. Similar triangles have the properties that their corresponding angles are the same (what I just said) and corresponding sides are in the same proportion.
For the larger right triangle, the base has length 40 and the height is 8. For the smaller triangle, the base is T and I will call the height "y". Those are in the same same proportion meaning the fractions formed are equal: $\frac{8}{40}= \frac{y}{T}$ and multiplying both sides by T gives $y= \frac{8}{40}T= \frac{1}{5}T$, not "5T". I had that wrong before! Sorry about that, I did it too fast.

Another, less "geometry", more "physics" method: The bus goes form 0 to 8 m/s in 40 s so it has a constant acceleration of 8/40= 1/5 m/s^2. At constant acceleration v= at so at t= T with acceleration 1/5, v= (1/5)T.

That is about the bus only and doesn't mention the bicycle so the fact that the bicycle goes twice as far as the bus is irrelevant.

The distance the bus has gone is the area of a triangle with base T and height (1/5)T which is (1/2)(T)(1/5)T= (1/10)T^2. The distance the bicycle goes in time T, at constant speed 5 is 5T (the area of a rectangle with base T and height 5). Since the bicycle travels twice as far as the bus, 5T= 2(1/10)T^2 or 5T= (1/5)T^2. Solve that for T. (One obvious answer is T= 0 since 2(0)= 0 but that is clearly not the solution you want. Since T is NOT 0, you can divide both sides by T.)