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help with theses would be overly appreciated i will rep one of the responders and the others to my rep list!

1) A Car is moving along a straight road with uniform acceleration. the ar passes check piunt a with a speed of 12ms^-1 and another checkepoint C with a speed of 32ms^-1. the distance between A and C is 1100m

find the time in seconds from a to c

given that b is the mid point of ac find in ms^-1 to 1 dec place the speed of which the car passes b

2) a b and c are three poins on a straight line in that order, the distances AB adn AC are 45m and 77m respectivly. a particle mpves along the straight line with a cnstant acceeration of 2ms^-2 / given that it takes 5 seconds to travel from a to b find time from a to c

3) a particle moves wit constant acceleration 0.5ms^-2 along a straight line passing through the points p and q. it passes the point q with a velocity of 1ms^-1 greater than the velocity at p. given the distance of pq is 25m calculate the velocity with which the particle passes the point p.

1) A Car is moving along a straight road with uniform acceleration. the ar passes check piunt a with a speed of 12ms^-1 and another checkepoint C with a speed of 32ms^-1. the distance between A and C is 1100m

find the time in seconds from a to c

given that b is the mid point of ac find in ms^-1 to 1 dec place the speed of which the car passes b

2) a b and c are three poins on a straight line in that order, the distances AB adn AC are 45m and 77m respectivly. a particle mpves along the straight line with a cnstant acceeration of 2ms^-2 / given that it takes 5 seconds to travel from a to b find time from a to c

3) a particle moves wit constant acceleration 0.5ms^-2 along a straight line passing through the points p and q. it passes the point q with a velocity of 1ms^-1 greater than the velocity at p. given the distance of pq is 25m calculate the velocity with which the particle passes the point p.

1) s = (t/2)(u+v)

So we have

v = 32

u = 12

s = 1100

rearranging;

t = (2s)/(u+v)

= 2200/44

= 50s.

______

Let b be the midpoint of a,c.

find the acceleration.

v=u+at

32=12+50a

50a = 20

a=20/50

a=2/5

now v^2 = u^2 + 2as

so v^2 = 12^2 + 2.(2/5).550

v^2 = 584

v = 24.166...

v = 24.2 to 1dp.

Sorry, i've had to rush! I'm going out in a second, and i've not done mechanics in years so someone tell me if i've got the equations wrong.

So we have

v = 32

u = 12

s = 1100

rearranging;

t = (2s)/(u+v)

= 2200/44

= 50s.

______

Let b be the midpoint of a,c.

find the acceleration.

v=u+at

32=12+50a

50a = 20

a=20/50

a=2/5

now v^2 = u^2 + 2as

so v^2 = 12^2 + 2.(2/5).550

v^2 = 584

v = 24.166...

v = 24.2 to 1dp.

Sorry, i've had to rush! I'm going out in a second, and i've not done mechanics in years so someone tell me if i've got the equations wrong.

(2)

A to B:

t= 5 s=45 a=2

s= ut + 0.5at²

45= 5u + 25 => u = 4

then from A to C

s= 77 a=2 u= 4

s= ut +0.5at²

77= 4t + t²

0= t² + 4t - 77

(t+2)² - 81 =0

(t+2)² = 81 => t= 7 seconds

A to B:

t= 5 s=45 a=2

s= ut + 0.5at²

45= 5u + 25 => u = 4

then from A to C

s= 77 a=2 u= 4

s= ut +0.5at²

77= 4t + t²

0= t² + 4t - 77

(t+2)² - 81 =0

(t+2)² = 81 => t= 7 seconds

(3)

call the velocity at p, u ms/1.

then from p to q:

a= 0.5 s= 25 u=u v=u+1

use

v² = u² + 2as to find u.

call the velocity at p, u ms/1.

then from p to q:

a= 0.5 s= 25 u=u v=u+1

use

v² = u² + 2as to find u.

silent ninja

(3)

call the velocity at p, u ms/1.

call the velocity at p, u ms/1.

?

Original post by Inspiron

1) s = (t/2)(u+v)

So we have

v = 32

u = 12

s = 1100

rearranging;

t = (2s)/(u+v)

= 2200/44

= 50s.

______

Let b be the midpoint of a,c.

find the acceleration.

v=u+at

32=12+50a

50a = 20

a=20/50

a=2/5

now v^2 = u^2 + 2as

so v^2 = 12^2 + 2.(2/5).550

v^2 = 584

v = 24.166...

v = 24.2 to 1dp.

Sorry, i've had to rush! I'm going out in a second, and i've not done mechanics in years so someone tell me if i've got the equations wrong.

So we have

v = 32

u = 12

s = 1100

rearranging;

t = (2s)/(u+v)

= 2200/44

= 50s.

______

Let b be the midpoint of a,c.

find the acceleration.

v=u+at

32=12+50a

50a = 20

a=20/50

a=2/5

now v^2 = u^2 + 2as

so v^2 = 12^2 + 2.(2/5).550

v^2 = 584

v = 24.166...

v = 24.2 to 1dp.

Sorry, i've had to rush! I'm going out in a second, and i've not done mechanics in years so someone tell me if i've got the equations wrong.

Why can't we just divide the time/2 as acceleration is constant?

Hit the thanks button if I helped!🙏

Original post by Yousef Khashaba

Why can't we just divide the time/2 as acceleration is constant?

Hit the thanks button if I helped!🙏

Hit the thanks button if I helped!🙏

That is only if the velocity is constant.

Imagine/sketch a displacement-time graph for a particle starting from the origin moving with:

(a) constant velocity

(b) constant acceleration

for a time T

For (a) the gradient/slope will be constant. For (b) the gradient/slope will increase.

Now draw a vertical line over when t=T/2.

You'll realise why the displacement wouldn't be half of the final displacement for (b) but it would be for (a).

Original post by AlphaNick

yeah just give him the answers ^_^

Well, this is a thread resurrected from 2006

Also the guy was asking for an explanation to clear up understanding, rather than the actual answer to the question.

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