Simple 2-D Uniform Motion Problem

[aishax]

A ball rolling with an initial velocity of 40 m/s [W] undergoes an acceleration of 5.0 m/s^2 [N] for a period of 6.0 seconds.

a) What is the final velocity of the ball?
b) What is the displacement of the ball in the 6.0 seconds?



Can someone explain how I would draw this vector diagram? What gets me stuck is how acceleration is in a different direction than the initial velocity. I have a test tomorrow on Kinematics, so all help is greatly appreciated! Thanks in advance :)

 

[venkatg]

V = U + a * t

S = U * t + (1/2) * a * t * t

V = final velocity
U = initial velocity
t = time taken
S = distance traveled in time t

 

[aishax]

Unfortunately, this problem isn't that simple.
The answer given is: a) 50 m/s [W37*N] b) 260 m [W21*N]

I assume it is required for me to draw a vector diagram, but the fact that this problem involves acceleration is what is confusing me.

 

[venkatg]

ok, I did not see the directions

Velocity in west direction = 40
velocity in N is U + a *t = 0 + 5 * 6 = 30

Now the final velocity is Northwest = sqrt ( 40^2 + 30 ^2) = 50 m/sec

azimuth (angle) from west = tan inverse(3/4) = 34 degrees W

Distance is sqrt [(X * X + Y *Y)]

X = 40 * 6 = 240 m
Y = 0.5 * 5 * 6 * 6 = 90

Distance = 256.3m

azimuth = tan inverse (90/240) = 20.55

 

[aishax]

Ahh, makes complete sense. Thanks so much!