1. Vectors for Mechanics

2. j i e.g. A velocity v is given by v  3  4 ) m s -1 i j x y j i 3 4 v Instead of drawing diagrams to show vectors we can use unit vectors. A unit vector has magnitude 1. The unit vectors and are parallel to the x- and y- axes respectively. ij In text-books single letters for vectors are printed in bold but we must underline them.

3. v  3  4 i j The magnitude of velocity is speed, so, using Pythagoras’ theorem, v  3 2  4 2 No or in magnitude i j v  3 2  4 2 So, if we have the unit vector form, we use the numbers in front of and ij x y 3 4 v j i Tip: Squares of real numbers are always positive so we never need any minus signs. v  5  We can write v or v for speed.

4. v   i j The direction of the vector is found by using trig. tan   53·1  ( 3 s.f. )  BUT beware ! 3443 x y 3 4 v j i

5. If we need the direction of a vector when unit vectors are used, we must sketch the vector to show the angle we have found. v  3  4 i j Suppose Without a diagram we get tan  33 44  53·1  ( 3 s.f. ) So again But, the vectors are not the same ! v   i j  53·1  ( 3 s.f. ) 3443 For we have 3 4  i j 3443 v   3 4 v  3  4 i j 

6. Equations of Motion for Constant Acceleration

7. We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t0 Suppose when the time is 0... At any time, t, we let the velocity be v. the velocity is u. Ans: The gradient gives the acceleration. Remind your partner how to find acceleration from a velocity-time graph. Constant acceleration means the graph is a straight line.

8. We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t a  v  u t So, v  u 0 Suppose when the time is 0... the velocity is u. t From this equation we can find the value of any of the 4 quantities if we know the other 3. At any time, t, we let the velocity be v. Constant acceleration means the graph is a straight line.

9. a  v  u t a t  v  u v  u  a t Multiplying by t :  u  a t  v  We usually learn the formula with v as the “subject”. The velocity, u, at the start of the time is often called the initial velocity.

10. Displacement and Velocity using Unit Vectors

11. e.g.A ship is at a point A given by the position vector r  A  4  3 ) km i j Solution: Find (a) the displacement of B from A, and After half-an-hour the ship is at a point B. We can solve this problem without a diagram, but a diagram can help us to see the method. i The ship has a constant velocity of 6 km h -1. (b) the position vector of B.

12. O y x 4 3 A x 6 i After half-an-hour the ship is at a point B. B x displacement  velocity  time  3 km i   s  6  i  4  3 ) i j  3 i    3 ) km i j r  A  4  3 ) km i j A:A: Velocity v  6 km h -1 i r B  r A  sr B  r A  s r B r B  r Br B r Ar A Solution: (a)Find the displacement of B from A. 0·5 (b)Find the position vector of B. Constant velocity  s