Acceleration due to gravity is -9.8m/s^2

[usagi101]

First Question:
Assume the acceleration due to gravity is -9.8m/s^2 and ignore air resistance. upward motion is POSITIVE direction

1.a) a stone is projected upwards from O with a speed of 20 m/s find the velocity of the stone after 4 s.

ok this is fairly easy so a=-9.8 u=20 t=4
v=u+at so v=20+(-9.8)4 v=-19.2m/s

the next part says repeat 1.a) for the stone being projected downwards from O with the same speed.

since its down then the acceleration is positive. so a=9.8 and the initial velocity (u) is negative since its DOWNWARDS from O. so u=-20 and t=4
v=u+at so v=-20+9.8x4 v=19.2m/s... however in my book it says that the acceleration is -9.8... and so then u sub that in the formula and u get v=-59.2 m/s. I am just not sure why the acceleration is negative when its downwards. Could some1 explain thanks!

Second Question:
A body is traveling at 20 m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q. assuming uniform acceleration.
ok this question i have no idea lol.. i tried some but i got nowhere. if anyone can explain step by step how to do it thanks!

Third Question:
A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
ok.. ill leave out some steps but through many checks the total distance(x) traveled is 500m so half of that is 250m. acceleration is worked out to be -5/18m/s^2 and initial speed (u) is 50/3 m/s.
subbing that in the formula --> x=ut+0.5at^2 250=(50/3)t+0.5(-5/18)t^2 and t is worked to be 17.6s or 102.43s but the answer in my book says its 21.96 s. Is it a mistake in the book or did i make a mistake? Thanks.

Thats all the questions. Thanks for anyone who can help!

 

[cristo]

First Question:
Assume the acceleration due to gravity is -9.8m/s^2 and ignore air resistance. upward motion is POSITIVE direction

1.a) a stone is projected upwards from O with a speed of 20 m/s find the velocity of the stone after 4 s.

ok this is fairly easy so a=-9.8 u=20 t=4
v=u+at so v=20+(-9.8)4 v=-19.2m/s

the next part says repeat 1.a) for the stone being projected downwards from O with the same speed.

since its down then the acceleration is positive. so a=9.8 and the initial velocity (u) is negative since its DOWNWARDS from O. so u=-20 and t=4
v=u+at so v=-20+9.8x4 v=19.2m/s... however in my book it says that the acceleration is -9.8... and so then u sub that in the formula and u get v=-59.2 m/s. I am just not sure why the acceleration is negative when its downwards. Could some1 explain thanks!

The book is correct. What you need to remember is to adopt sign conventions. So, in your conventions, upwards is positive. Since the velocity is downwards, it will have a negative sign, and since the acceleration is also downwards, it will also have a negative sign. Of course, since these are just conventions, the important thing is that they both have the same sign.

Second Question:
A body is traveling at 20 m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q. assuming uniform acceleration.
ok this question i have no idea lol.. i tried some but i got nowhere. if anyone can explain step by step how to do it thanks!

What did you try? The key here is uniform acceleration.

Third Question:
A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
ok.. ill leave out some steps but through many checks the total distance(x) traveled is 500m so half of that is 250m. acceleration is worked out to be -5/18m/s^2 and initial speed (u) is 50/3 m/s.
subbing that in the formula --> x=ut+0.5at^2 250=(50/3)t+0.5(-5/18)t^2 and t is worked to be 17.6s or 102.43s but the answer in my book says its 21.96 s. Is it a mistake in the book or did i make a mistake? Thanks.

Thats all the questions. Thanks for anyone who can help!

I can't check this until you put your steps in: so, for example, where fors -5.18m/s^2 come from?

 

[tiny-tim]

Hi usagi101!

A body is traveling at 20 m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q. assuming uniform acceleration.

erm … this is easy … think!

A tram decelerates uniformly from a speed of 60km/h to rest in 60 s find the time taken for it to travel half the total distance.
ok.. ill leave out some steps but through many checks the total distance(x) traveled is 500m so half of that is 250m. acceleration is worked out to be -5/18m/s^2 and initial speed (u) is 50/3 m/s.
subbing that in the formula --> x=ut+0.5at^2 250=(50/3)t+0.5(-5/18)t^2 and t is worked to be 17.6s or 102.43s but the answer in my book says its 21.96 s. Is it a mistake in the book or did i make a mistake? Thanks.

i] It would be a lot easier if you used units of minutes and km/minute, wouldn't it?
ii] You don't need to solve the equations … just use dimensions … what is t proportionate to?
iii] I think the book answer is wrong.