Question

Two bodies more from the same point along straight line.the first body moves
with velocity v = (3t2 - 6t) m/s,the second with velocity v = (10t+20)m/s.At
what instant and at what distance from the initial point will they meet​

Answer 1

Given: Two bodies more from the same point along straight line.the first body moves  with velocity v = (3t^2 - 6t) m/s,the second with velocity v = (10t+20)m/s.

To find: At  what instant and at what distance from the initial point will they meet​?

Solution:

• Let v1 = (3t^2 - 6t) m/s and v2 = (10t+20)m/s

• Now both bodies will meet when distance travelled will be equal, so:

                 dx1 /dt = (3t^2 - 6t)  and  dx2 / dt = (10t+20)

                 x1 = 3t^3/3 - 6t^2/2 and x2 = 10t^2/2 + 20t

                 x1 = t^3  - 3t^2 and x2 = 5t^2 + 20t

• So, x1 = x2

                 t^3  - 3t^2 = 5t^2 + 20t

                 t^3  - 8t^2 - 20t = 0

                 t x (t^2 - 8t - 20) = 0

                 t = 8 ± √64 - 4(-20) / 2

                 t = 8 ± √64+80 / 2

                 t = 8 ± √144 / 2

                 t = 8 ± 12 / 2

                 t = 20/2 , -4/2 , 0

                 t = 10, -2 , 0

                 -2 will be discarded.

Answer:

            So t = 0 or t = 10 second.

Answer 2

Answer:

t = 10 second..is the answer

Explanation: