ashoni's question thread! :)

[ashoni]

Would appreciate it if someone could help me with this question

A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10s and reaching a constant speed of 100 km/h after a further 5s. At what time will the policeman catch up with the car?

[kaiipoo_]

imagine a graph of velocity on time.

the speed of the car would be (80/3.6) = 22.22 m/s
and therefore, distance travelled would = 22.22t given by d=s.t

On a graph of v vs t, the car's motion would appear to be a straight line, as he is travelling at a constant speed.

For the Policeman, it's a little different, as his motion changes from a sharp acceleration -> smaller acceleration -> constant speed

So, we know that on a velocity vs time graph, the area under the graph gives the distance. To make it easier we should break it up.

- for the first 10 s, acceleration is uniform and thus you'll have a straight line with a positive gradient on the graph.

for the distance travelled, using the area of a triangle equation A = 1/2bh, A= 1/2x10x22.22 --> giving us d= 111.11 m

- for the time interval t=10 -> t=15, the policeman reaches a final speed of 100 km/h, which is 27.78 m/s. This part of the graph gives a one sided trapezium thingy lol, and therefore, the distance travelled in this interval would be:

A(triangle) + A(rectangle) (after breaking up the shape)

A(triangle) = 1/2 x 5 x (27.78-22.22) = 13.9m
A(rectangle) = 22.22 x 5 = 111.1 m

and therefore, distance travelled = 125 meters.

so, the entire distance travelled in the ENTIRE 15s = 125+111 = 236.1 meters.


Now, we need to find the distance travelled by the car, which is 22.22x15 = 333.3 meters.

Therefore, the cop travels 236, and the car travels 333, so they are (333.3-236.1) meters apart, which = 97.2 m

so now we arrive at the last step, now we need to find how long it takes for the cop to catch up with the car, now both travelling at constant speeds.

the car's speed = 22.22 m/s
the motorcycle's speed = 27.78 m/s (5.56 m/s faster than the car)

therfore, t= d/s
t= 97.2/5.56
t= 17.5 seconds.

add that to the initial 15 seconds, and you should get your final answer of:

t= 32.5 s

[atar92point4]

80km/h = 22.22m/s
for the police to catch the car they must have travel same distance hence displacement of both equal

So after t seconds the car travels 22.22t m  (v=d/t and d=vt)
But the police (first 10s ) has displacement 111m using (x=ut+0.5at^2)
Police (next 5s) has displacement of 125m again using (x=ut+0.5at^2)

Now after t seconds the police has travelled   x=27.77*(t-15) + (125+111)  [T= t-15 because we calculated the displacement in 15s]

again as i said for the police to catch the car they must have travel same distance hence displacement of both equal
so 22.22t=27.77*(t-15) + 236
22.22t=27.77t - 416.66+236
22.22t=27.77t-180.66
-5.55t=-180.66
therefore t=32.5s

[ashoni]

Thank you guys for your responses! Much appreciated!

[ashoni]

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot of the table. This starship now lands on the unexpected Planet X. The commander, Captain Curious rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76D from the foot of the table. What is the acceleration due to gravity on planet X?

need help with the explanation to this question...

[availn]

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot of the table. This starship now lands on the unexpected Planet X. The commander, Captain Curious rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76D from the foot of the table. What is the acceleration due to gravity on planet X?

need help with the explanation to this question...

First, recognise that the horizontal velocity is constant and the same at both planets. Therefore, it takes 2.76 times as long for the ball to hit the ground on Planet X than it does on Earth.

We'll use the formula s = ut + 0.5at², as that equation does not use final velocity, which we don't know. The table is the same height on both planets (s), and the initial vertical velocity (u) is 0 (horizontal velocity and distance is now unneeded, it was in the question so we could find the times for the ball to fall). So:

0.5gt² = 0.5at²

Where g is Earth's acceleration due to gravity, and a is Planet X's acceleration due to gravity. We can sub in 1 and 2.76 for the two values of t.

5 = 3.81a
a = 1.31 ms^-2

We can check if this is correct by working it backwards, with a table of, say, 1 metre height, and by finding the time it takes for the ball to fall on both planets. On earth:

s = ut + 0.5at²
1 = 5t²
t = 0.45 seconds

On Planet X:

s = ut + 0.5at²
1 = 0.655t²
t = 1.24 seconds

1.24 is 2.76 times larger than 0.45, so the ball would travel 2.76 times further horizontally on Planet X.

[ashoni]

First, recognise that the horizontal velocity is constant and the same at both planets. Therefore, it takes 2.76 times as long for the ball to hit the ground on Planet X than it does on Earth.

We'll use the formula s = ut + 0.5at², as that equation does not use final velocity, which we don't know. The table is the same height on both planets (s), and the initial vertical velocity (u) is 0 (horizontal velocity and distance is now unneeded, it was in the question so we could find the times for the ball to fall). So:

0.5gt² = 0.5at²

Where g is Earth's acceleration due to gravity, and a is Planet X's acceleration due to gravity. We can sub in 1 and 2.76 for the two values of t.

5 = 3.81a
a = 1.31 ms^-2

We can check if this is correct by working it backwards, with a table of, say, 1 metre height, and by finding the time it takes for the ball to fall on both planets. On earth:

s = ut + 0.5at²
1 = 5t²
t = 0.45 seconds

On Planet X:

s = ut + 0.5at²
1 = 0.655t²
t = 1.24 seconds

1.24 is 2.76 times larger than 0.45, so the ball would travel 2.76 times further horizontally on Planet X.

ohhh that makes sense!! thanks for your help