Motion Questions . Argh stuck!

[squance]

My sister is stuck on two questions from her physics book.

A car travelling with a constant speed of 80 km h^-1 passes a stationary policeman . The policeman sets off in pursuit, accelerating uniformly to 80 km h^-1 in 10.0 s and reaching a constant speed of 100 km h^-1  after a further 5.0 s. At what time will the policeman catch up with the car?


The answer is supposedly 33 s but my sister can't get it (well...she actually has no clue on how to do the question).


The other question is:

A satellite is in a geosynchronous orbit around the Earth if its period of rotation is the same as that of the Earth, i.e 24h. Such a satellite is called a geostationary satellite. Venus has a mass of 4.87 x 10^24 kg and a radius of 6.05 x 10^6 m. The length of day on Venus is 2.10 x 10^7 s. For a satellite to be in a synchronous orbit around Venus, calculate:
a). The orbital radius of the satellite (actually, my sis has already worked out this part which is R = 1.54 x 10^9 m.
b). Its orbital speed
c). Its orbital acceleration

She can't do b or c

Any help would be greatly appreciated since I have no knowledge on Physics

[dekoyl]

80km/h = 22.22m/s.
Draw a graph for this (v vs t).

Area under graph = distance.
Car: 22.22t
Policeman: .5(10)(22.22) + 32(t - 10)
22.22t = 0.5(10)(22.22) + 32(t-10)
Then solve for t.

Sorry it's rushed I have to go to dinner. If she doesn't understand I'll explain after it.

[danieltennis]

where , M = mass of Venus,

c/ Sub the value of v from the previous answer and R = 1.54 x 10^9

[squance]

Hey. Thanks for the responses daniel and dekoyl.

80km/h = 22.22m/s.
Draw a graph for this (v vs t).

Area under graph = distance.
Car: 22.22t
Policeman: .5(10)(22.22) + 32(t - 10)
22.22t = 0.5(10)(22.22) + 32(t-10)
Then solve for t.

Sorry it's rushed I have to go to dinner. If she doesn't understand I'll explain after it.

Dekoyl, my sis doesn't understand how you have 32(t-10) in your working.

[dekoyl]

Do I get the right answer?
Anyway, on the graph for the police, it would look like
(v)
|  ________
|/_________>(t)

See the shift from diagonal to horizontal? The 0.5(10 seconds)(22.5 m/s) is the distance travelled during acceleration.
Then, 32(t - 10) is the distance travelled during constant speed. However, I do t - 10 because 10 seconds has already been accounted for when calculating the distance travelled during acceleration. Hence I can't use the 10 seconds when calculating the area under the horizontal part.

[squance]

Im looking at my sister's piece of paper and she hasn't worked it out yet.
But when I solved that equation myself, I got 21.24 :S :S

[dekoyl]

Oh my fault. I didn't see  the reaching bit. Sorry was in a rush. I'll re do it.

[TrueTears]

the area under neath a velocity - time graph gives the distance.

lol sorry, i forgot the change in gradient ( in the graph between 10 and 15 seconds that line shud have a different gradient) haha, dekoyl's working is correct.
v

v

v


EDIT: the speeds are given in make sure you convert to by dividing 3.6 (when typing on calc use the fraction to get a more accurate answer)

[dekoyl]

Keeping TT's graph in mind,
After 10 seconds, the acceleration isn't the same (I think) so the gradient changes again (hence you have to recalculate the area between t = 10 and 15).



Solving will yield

[Damo17]

I did this question here:
http://vcenotes.com/forum/index.php/topic,8915.msg114304.html#msg114304

[squance]

Hehe. Thank you everyone for helping.
Damo17, my sis understood your working the clearest.