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Mechanics- Kinematics Help

Hi everyone, just wondering if you could help me with some of my mechanics. Heres some of the questions:

q.1) A car starts from rest at a point O and moves in a straight line. The car moves with constant acceleration 5ms(-2) until it passes the point A when it is moving with speed 5ms(-1). It then moves with constant acceleration 3ms(-2) for 7 s until it reaches the point B. Find

a) the speed of the car at B
b) the distance OB


q. 2) A train T1, moves from rest at station A with constant acceleration 2ms(-2) until it reaches a speed of s=28m/s. It maintains this constant speed for t=60s before the brakes are applied, which produce constant retardation 2ms (-2)

a) Show that the distance betwen A and B is 2072m.
b) Find the greatest speed, in m/s, attained by T2 during its journey.

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q. 3) An aircraft moves along a straight horizontal runway with constant acceleration. It passes at point A on the runway with speed 14m/s. It then passes the point B on the runway with speed 30m/s. The distance from A to B is 160m.

a) Find the acceleration of the aircraft.
b) Find the time taken by the aircraft in moving from A to B.
c) Find the speed of the aircraft when it passes the point mid-way between A and B.
Reply 1
1a) v = u + at

v = 5 + 3*7 = 26 m/s

b) s = ut + 0.5at²

the first section must take 1 second ( 5 = 0 + 5*t )

the first distance must be 0*1 + 0.5*5*1²

the second section takes 7 seconds ...

s = 5*7 + 0.5*3*7²
Reply 2
q.4) A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 10m/s. It passes the point A 5s later with speed 53m/s.

a) Show that the acceleration of the car is 8.6ms(-2)
b) Find the distance OA
The point B is the mid-point of OA
c) Find the speed of the car when it passes point B.


q.5) A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 12m/s. It passes the point A 2s later with speed 60m/s.

a) Show that the acceleration of the car is 24ms(-2)
b) Find the distance OA
The point B is the mid-point of OA
c) Find the speed of the car when it passes point B.


q.6) A racing car moves along a straight horizontal road with constant acceleration. The car passes the point A with speed 4ms(-1) and 5 s later it passes point B, where AB=52M

a) Find the acceleration of the car
When the car passes the point C, it has speed 28ms(-1)
b) Find the distance AC

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all help is much appreciated...............thanks bear.
Reply 3
Bowie88
q.4) A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 10m/s. It passes the point A 5s later with speed 53m/s.

a) Show that the acceleration of the car is 8.6ms(-2)
b) Find the distance OA
The point B is the mid-point of OA
c) Find the speed of the car when it passes point B.


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all help is much appreciated...............thanks bear.


a) acceleration = change in speed/time = 43/5 = 8.6 m/s²

b) = + 2as

53²= 10² + 2*8.6*s

so s = (53² - 10²)/(2*8.6)

c) use = + 2as with u = 10 but with s = half of the answer to b)
Reply 4
the bear
a) acceleration = change in speed/time = 43/5 = 8.6 m/s²

b) = + 2as

53²= 10² + 2*8.6*s

so s = (53² - 10²)/(2*8.6)

c) use = + 2as with u = 10 but with s = half of the answer to b)


I had nothing better to do so I did this one, to find Bear already explained most of it , here it is anyway.
(a) v=u+at
.....a=(v-u)/t
.....a=(53-10)/5
.....a=43/5 = 8.6m/s

(b) =u² +2as
.....s=(v²-u²) /2a
.....s=(53² -10²) /(2x8.6)
.....s=2709/17.2 = 157.5m

(c) Distance OB= 157.5/2 = 78.75m
.....v² =u² +2as
.....v² =10² +(2x8.6x78.75)
.....v² = 1454.5
.....v=38.1m/s (1dp)
Reply 5
thanks both of you, i will send my rep to you both. Could you help me with the rest of the questions please?
Reply 6
Bowie88

q.5) A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 12m/s. It passes the point A 2s later with speed 60m/s.

a) Show that the acceleration of the car is 24ms(-2)
b) Find the distance OA
The point B is the mid-point of OA
c) Find the speed of the car when it passes point B.



This is more or less same then Question 4:
Q5
(a) v=u+at
.....a=(v-u)/t
.....a=(60-12)/2
.....a=48/2 = 24m/s²

(b) =u² +2as
.....s=(v²-u²) /2a
.....s=(60² -12²) /(2x24)
.....s=3456/48 = 72m

(c) Distance OB= 72/2 = 36m
.....v² =u² +2as
.....v² =12² +(2x24x36)
.....v² = 1872
.....v=43.3m/s (1dp)
Reply 7
Bowie88


q.6) A racing car moves along a straight horizontal road with constant acceleration. The car passes the point A with speed 4ms(-1) and 5 s later it passes point B, where AB=52M

a) Find the acceleration of the car
When the car passes the point C, it has speed 28ms(-1)
b) Find the distance AC

6QA
s=ut+0.5(at²)
52=(4x5)+12.5a
12.5a=52-20
12.5a=32
a=2.56m/s²

(B)
v²=u²+2as
s=(v²-u²)/2a
s=(28²-4²)/5.12
s=768/5.12 =150m
Reply 8
thanks mala2k................sorry but ive got a couple more questions!

q.7 A car accelerates uniformly from rest to a speed of 30ms(-1) in T seconds. The car then travels at a constant speed of 30ms(-1) for 4T seconds and finally decelerates uniformly to rest in a further 60s.
The total distance travelled by the car is 1170m. Find

a) the value of T
b) the initial acceleration of the car


q.8 A competitor makes a dive from a highboard into a diving pool. She leaves the board vertically with a speed of 3m/s upwardds. When she leaves the board she is 5m above the surface of the pool. The diver is modelled as a particle moving vertically under gravity alone and it is assumed that she does not hit the springboard as she descends. Find

a) her speed when she reaches the surface of the pool
b) the time taken to reach the surface of the pool
c) State two physical characteristics which have been ignored in the model.
Reply 9
Bowie stop typing more questions!

We won't do them all. Try em' yourself.
Reply 10
Ricki
Bowie stop typing more questions!

We won't do them all. Try em' yourself.


whatever!
Bowie88
whatever!

god, you got 12A*?
Try some yourself!
Reply 12
Widowmaker
god, you got 12A*?
Try some yourself!


that was gcses! They were so easy compared to alevels. I know, i know you are all still wondering why im asking these questions, the only reason i can't do these is because i have been spending too much time on my alevel art, but i thought you all could help me with mechanics.
Reply 13
a little help here people!
Bowie88
a little help here people!

you obviously don't understand the basics of this topic. I'd speak to your teacher. If you have the m1 book, read through the examples and the theory as it really helped me when I did M1. I don't really think we should help you because tbh you aren't learning anything.
Reply 15
Widowmaker
you obviously don't understand the basics of this topic. I'd speak to your teacher. If you have the m1 book, read through the examples and the theory as it really helped me when I did M1. I don't really think we should help you because tbh you aren't learning anything.


ok, thanks for the advice, i just wanted to know how to do these questions, could you help me?
Reply 16
please, this is a last plea for help...................please!!!!!!!!!!!
Bowie88
please, this is a last plea for help...................please!!!!!!!!!!!

no offence mate, but some of us have our own homework to do. :rolleyes:
Reply 18
Widowmaker
no offence mate, but some of us have our own homework to do. :rolleyes:


ok............point taken, im sorry for wasting your time
Reply 19
Bowie88
Hi everyone, just wondering if you could help me with some of my mechanics. Heres some of the questions:

q. 2) A train T1, moves from rest at station A with constant acceleration 2ms(-2) until it reaches a speed of s=28m/s. It maintains this constant speed for t=60s before the brakes are applied, which produce constant retardation 2ms (-2)

a) Show that the distance betwen A and B is 2072m.
b) Find the greatest speed, in m/s, attained by T2 during its journey.



a) if you plot the speed time graph you get a trapezium. the distance travelled is the area of it. the left hand triangle has area t*28/2
the acceleration =28/t so t=28/2 = 14 sec. the distance is 14*28/2 metres = 196 metres
the middle section is a rectangle with area 60*28
the last triangle is congruent to the first triangle so has area 196 metres
so the total distance is 196 + 1680 + 196 = 2072 m