Question

A particle moving along a straight line with
uniform acceleration has velocities 7 ms-1 at A

and 17 ms1 at B. C is the midpoint of AB. The
ratio of time taken to go from A to C and that

from C to B is

2:3

4:3

3:2

3:4



​

Answer 1

Explanation:

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = s

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2as

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor AC

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2as

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC289 - u^2=2as -(2)

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC289 - u^2=2as -(2)from equation 1 and 2, we get

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC289 - u^2=2as -(2)from equation 1 and 2, we getu^2–49=289- u^2

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC289 - u^2=2as -(2)from equation 1 and 2, we getu^2–49=289- u^2on solving we get

Here you have not mentioned anything about acceleration . so i making this question by assuming that acceleration is uniform.let u be the velocity at B.Since B is the mid point of AC.so, AB = BC = sso ,on applying the formula v^2- u^2 = 2asfor ACu^2- 7^2= 2as - (1)u^2–49= 2assimilarly for BC289 - u^2=2as -(2)from equation 1 and 2, we getu^2–49=289- u^2on solving we getu= 13 m/s