Question

Two particles A and B are moving in the same direction on parallel horizontal tracks. At a a certain point the particle A travelling with a speed of 7ms^-1 and accelerating uniformly at 1.5ms^-2 overtakes B travelling at 3ms^-1 and accelerating uniformly at 2.5ms^-2. Calculate the period of time which elapses before B overtakes A. If, after this time, B then ceases to accelerate and continues at constant speed, calculate the rime taken for A to overtake B again.

Answer 1

i answered the first part

Answer 2

did you get 8s?

Answer 3

s=7t+1/2(1.5)t^2=3t+1/2(2.5)t^2, 7t+3/4t^2=3t+5/4t^2. we get rid of denominators by multiplying the 4 so 28+3t^2=12t+5t^2, 2t^2-16t solve quadratic equation t=8

Answer 4

i cant figure out the 2nd part of the question

Answer 5

it's the same as the first part. You just need to know how fast each particle is going when they're side by side.

Answer 6

what's is the speed of each particle after 8 seconds?

Answer 7

i dont have the speed of particle B

Answer 8

and the acceleration too!

Answer 9

ok, what's the speed of particle A then?

Answer 10

A has a speed 3ms

Answer 11

no

Answer 12

can you please read the question

Answer 13

i cant understand the question well

Answer 14

both particles accelerate, specifically for 8s, their speed will be greater initially yes?

final speed = (acceleration) (time) + initial speed

A = (1.5)(8) + 7
B = (2.5)(8) + 3

Answer 15

B ceases to accelerate, so no it's going at the speed of 23m/s. But A keeps accelerating at 1.5 m/s^2

A = 0.5(1.5)t^2 + 19t
B = 23t

solve

Answer 16

could you please explain the question to me iam a bit confused

Answer 17

it's the same equation you used in part 1. Just with different numbers

Answer 18

s = (1/2)(acceleration) (time) + (initial velocity) (time)

Answer 19

okay i understand this part but how did you figure out these 2 equations A = 0.5(1.5)t^2 + 19t
B = 23