A body is moving in a straight line with constant acceleration one second after passing point A. It is 10m from A in the next second it travels a further 14m. Find the velocity of passing A and acceleration on passing through point A.​



Answer :

Answer:

u = 8 m/s

a = 4 m/s²

Explanation:

The body moves at constant acceleration. Using kinematic equations, also known as SUVAT, we can use the data to write a system of equation. The kinematic equation we'll use is:

s = ut + ½ at²

where

  • s is displacement (the final position relative to the initial position)
  • u is the initial velocity
  • a is the acceleration
  • t is time

After 1 second, the body is 10 m past point A.

10 = u (1) + ½ a (1)²

10 = u + ½ a

After 2 seconds, the body is 10 + 14 = 24 m past point A.

24 = u (2) + ½ a (2)²

24 = 2u + 2a

Solve the system of equations. Double the first equation and subtract from the second.

24 − 2(10) = 2u + 2a − 2(u + ½ a)

4 = a

a = 4 m/s²

Substituting into the second equation:

24 = 2u + 2 (4)

u = 8 m/s