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[Arithela]

1. A particle has constant acceleration of 12m/s^2. If the particle has a velocity of 2m/s and is 3m from the origin after 5s, find its displacement after 10s.

2. A projectile is accelerating at a constant rate of -9.8m/s^2. If it is intially 2m high and has a velocity of 4m/s, find the equation of the height of the projectile.

 

[vds700]

1. A particle has constant acceleration of 12m/s^2. If the particle has a velocity of 2m/s and is 3m from the origin after 5s, find its displacement after 10s.

2. A projectile is accelerating at a constant rate of -9.8m/s^2. If it is intially 2m high and has a velocity of 4m/s, find the equation of the height of the projectile.

1 dv/dt = 12
v = Int 12dt
=12t + c
v = 2 when t = 5
2 = 60 + c
therefore c = -58
therefore v = 12t - 58
x = Int(12t - 58)dt
=6t^2 -58t + k
x = 3 when t = 5
3 =150 -290 + k
k = 143
therefore x = 6t^2 - 58t + 143
when t = 10, x = 600 - 580 +143= 163 m.

 

[vds700]

13m @ 10seconds I believe.

did i make a mistake? Show me where. Sorry i did it straight on the computer, should have written it out on paper first.

 

[vds700]

ok i fixed up a mistake, but still get a different answer.

2.
y: = -9.8
y. = -9.8t + c
when t = 0, y. = 4sinA where A is the angle of prjection., therefore c = 4sinA
therefore y. = 4sinA - 9.8t
y = 4tsinA - (9.8t^2)/2 + k
when t = 0, y = 2, therefore k = 2
y = 4tsinA - (9.8t^2)/2 + 2 which is the equation of the height