Ineed !
1) since vectors are defined by their direction and magnitude, what does having multiple starting and ending points to create the same vector tell you about how you can manipulate vectors on graphs?

2) prove that the resultant vector can be found by the method of adding vectors described above. how did you prove this?

applying head to tail rule

Explanation:

1) As vectors are defined by their direction and magnitude it is very easy to create the vector having the multiple starting and ending points by the help of head to tail rule. in this method we choose a scale of units suitable to draw on graph (for instance 1kg may represent 1cm ) so that the head of the first vector coincides with the tail of the next one . so we get the resultant vector be drawing a vector in such a way that the tail of the first vector coincides with the head of the last one. the resultant vector will be drawn in such a way that its tail coincides with the tail of the first vector and head with the head of the last vector.

2) suppose we have 3 vectors having 3,4,5 units of magnitude and directed from south to north.

we will draw these along the y -axis on the same line staring with the first , then the second from the head of the first and then the third starting with the head of the second. we will get the resultant by drawing a vector starting from the tail of the first to the head of the last . this will give the resultant vector showing units of magnitude (3+4+5=12) and direction (from south to north).

the answer is a. hope this .

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s