Two particles, A and B, are initially 37.5 m apart. A is projected in a straight line directly towards B with initial speed 2 ms^–1 and accelerates uniformly at 0.6 ms^–2. At the instant when A is released, B is projected in a straight line directly towards A with an initial speed of 3 ms^–1 and it accelerates uniformly at 0.4 ms^–2. The particles collide at point C. Find the distance from A's starting point to C. (Please explain how you get the answer, I don't understand how to get to it Thanks!)

6.66 metres

Step-by-step explanation:

Given that the two particles, A and B, are initially 37.5 m apart.

Acceleration a = velocity/time

Make time t the subject of the formula.

Time t = velocity/acceleration

Time taken by particle A will be

Time t = 2/0.6 = 3.33 seconds

Let's use 2nd equation of motion

S = Ut + 1/2at^2

Distance covered by particle A will be achieved By substitute the values for speed and acceleration into the formula

S = 2 × 3.33 + 1/2 × 0.6 × 3.33^2

S = 6.66 + 0.3 × 11.1

S = 9.99 m

Time taken by particle B will be

Time t = 3/0.4 = 7.5 seconds

Distance covered by particle B will be by using the same formula

S = 3 × 7.5 + 1/2 × 0.4 × 7.5^2

S = 22.5 + 0.2 × 56.25

S = 33.75 metres

Adding the distance covered by particle A

Distance = speed × time

Distance = 2 × 3.33 = 6.66

the distance from A's starting point to C. Is 6.66 metres

answer: the answer is 40.82

step-by-step explanation: 3.14 x 13(or half of 26) to the power of 2