Question

Please help :)

Answer 1

Two particle are travelling along a straight line AB of length 20m. At the same instant one particle starts from rest at A and travels towards B with a constant acceleration of 2ms^-2 and the other particle starts from rest at B and travels towards A with a constant acceleration of 5ms^-2. Find how far from A the paricles collide.

Answer 2

@ash2326 @radar @AccessDenied @dpaInc @Calcmathlete

Answer 3

here, time travelled is same for both the particles when they collide

Answer 4

lets say, A travels x meters toward B with constant acceleration 2ms^-2
then B travels (20-x) meters toward A with constant acceleration -5ms^-2

Answer 5

how can the time be same as long as acceleration is diffeeent>

Answer 6

\[x = \frac{1}{2}2*t^2\]

Answer 7

\[20-x = \frac{1}{2}(-5)*t^2\]

Answer 8

i did it, can u check if i am right

Answer 9

time is same because, they started at same instant of time

Answer 10

how much u got ?

Answer 11

5.72

Answer 12

m

Answer 13

yup... i got the same :)

Answer 14

this is how i did...can u check if my method is corrct

Answer 15

ya.. sure post

Answer 16

i used the equation = v = u + at to find the final velocity for both
i got first one as v = 2t and the other one as v = 5t

i then used the equation displacement = vt - 1/2at^2
and substituted v and formed a simultaneous equation as below

so
20-s = 2t(t)- 1/2at^2
20-s = 2t^2 - t^2
20-s = t^2 -----> (1)

20-d = 5t(t) - 1/2at^2
20-d = 5t^2 - 5/2t^2
20-d = 2.5t^2

20= t^2 + s
20 = t^2 +d

i found and got
30 = 2.5s - d
30 = 2.5s - (20-s)
30 = 2.5s - 20 +s
50 = 3.5s
s = 14.29
d= 20 - 14.29
d = 5.71