Can any kind soul help me

a) 4.50m/s²

b) v= 21

c) 11.25m/s

Step-by-step explanation:

a) rate of change of speed in first 2 seconds

= gradient of graph from t=0s to t=2s

The two points are (0,0) and (2,9).

Rate of change of speed

Total distance= area under graph

Distance travelled for first 6s

= area of traingle +area of rectangle

= ½(2)(9) +(6-2)(9)

= 9 + 4(9)

= 9 +36

= 45m

Distance travelled from t=6s to t=12s

= 135 -45

= 90m

Area under graph from t=6s to t=12s =90

Area of trapezium= 90

½(A +B)(h)= 90

½(9+ v)(6)= 90

3(9 +v)= 90 (simplify)

9 +v= 30 (÷3 throughout)

v= 30 -9

v= 21

c) Average speed

= total distance ÷total time

= 135 ÷12

= 11.25 m/s

answer: i believe its b. if you sound out the sentences that sounds wrong.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this