velocity ,acceleration and projectile?

Question

1.A car's velocity as a function of time is given by Vsubscript of x(t)=alpha+beta*t^2,where alpha=3m/s & beta=0.100m/s^3.a.)calculate the average acceleration for the time interval t= 0 to t=5 sec.b.)calculate the instanteneous acceleration for i.)t=0;ii.)t=5 sec

2.A dog running in an open field has components of velocity Vsubscript x=2.6 m/s and Vsubscript y=-1.8 m/s at tsubscript 1 =10 s .For the time interval from t subscript 1 =10 s to t subscript 2 = 20 s,the average acceleration of the dog has magnitude 0.45 m/s^2 and the direction 31 degree measured from the +x-axis toward + y-axis. At tsubscript 2 =20 s, a.)What are the x and y components of the dog's velocity? b.)what are the magnitude and direction of the dog's velocity

3. Jillian angrily throws her engagement ring staright up from the roof of a building,12 m above the ground, with an initial speed of 5 m/s.you may ignore air resistance.For the motion from herhand to the ground,what are the magnitude & direction of a.)the average velocity of the ring? b.)the average acceletarion of the ring? c.)in how many seconds after being thrown does the ring strike the ground? d.) what is the speed of the ring just before it strikes the ground?

Answer 1

questions are very easy. You must try them by yourself. I will only give hints-
1> Instantaneous accl = dv/dt
Avg accl = (v2-v1)/(t2-t1)

2> draw the vectors, take the components separately along x and y axes.
remember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.
3> use the basic kinematics equations like v^2=u^2+2as etc

Answer 2

I have no idea what you are trying to ask so lets "unconfuse" the situation. When you throw a ball, the ball starts at rest, you apply a force which accelerates the ball and then you let go and the ball has a velocity but YOU can't accelerate the ball once it's left your hand. After that gravity and air resistance and other things can accelerate it (note that acceleration is just the change in velocity). So if it leaves your hand with only a velocity in the x direction, then with no acceleration by anything then the velocity will never change, in the x or y directions. But if there is gravity that is only acting in the y direction, then the ball will always have the same velocity in the x direction (because no x forces/acceleration) and will increase in velocity in the y direction because of the constant force/acceleration of gravity. Think about this and then repost a more clear question. Also, when you repost, use throw or threw instead of through.