Help with science: motion, velocity and acceleration?

Question

I'm stuck, i need someone to show me how to do this question(not just do it for me):

2. A particle, initially at rest, moves with accelerations of 3.00 m/s^2 due north and 4.00m/s^2 due east. These are maintained for 5.00 s. What is:

a) the velocity of the particle at the end of the interval;

b) the average speed;

c) the displacement?

Answer 1

OK.

Treat the north and east components separately, then do a vector addition on them to give the answer:

Going North.  v = u + at

v = 0 + 3 * 5 = 15 m/s    (Vnorth = 15 m/s)

Going East  v = u + at

v = 0 + 4 * 5 = 20 m/s  (Veast = 20 m/s)

Drawing a diagram is always a good idea!  So now you have two velocity vectors that you need to find the resultant for.

Resultant = sqrt( (Vnorth)^2 + (Veast)^2)  = sqrt( 225 + 400) = 25 m/s

If the question just asks for the MAGNITUDE of the velocity, then you're done.  In this case it asks for the velocity, and velocity is a vector which has both magnitude and direction, so we need to work out the angle or compass bearing as well.

Angle = arctan(4/3) = 53.13 degrees (east of north)

Do the same for displacement:

s = ut + 0.5at^2

North

s = 0 * 5 + 0.5 * 3 * 25 = 37.5 m   (Snorth = 37.5 m)

East

s = 0 * 5 + 0.5 * 4 * 25 = 50 m (Seast = 50 m)

Resultant displacement = sqrt(1406.25 + 2500) = 62.5 m

(Same compass bearing as the first part)

The average speed is displacement/time = 62.5/5 = 12.5 m/s

[You could also have worked this out using (u + v)/2 = 25/2 = 12.5]

In this particular question, they have given you "friendly" values for the vector components in that they are based on a 3-4-5 triangle

You can see that the resultant acceleration vector = 5 m/s^2 directly, so you could use v = u +at on that to give v = 0 + 5 * 5 = 25 m/s directly as the velocity, and the same for s.  But you need to be able to do the thing for any values!

HTH

Answer 2

v_t = v_0 + at = 0 + 7 x 5 = 35 m/s

average speed (v_av)= (v_0 + v_t)/2 = 35/2 = 17,5 m/s

t (while moving to north) = v_av/a_1 = 17,5/3=5,8 s

t (while moving to east) = v_av/a_2 = 17,5/4=4,375 s

distance crossed to north = 17,5x5,8=101,5 m

distance crossed to east = 17,5x4,375=76,5625 m

the displacement =sqrt [(distance crossed to north)^2 + (distance crossed to east)^2] = 127.14 m ≈ 127 m