Question

An object uniformily accelerates from 15.0m/s west to 35.0 m/s west. What is the magnitude of acceleration if the displacement during this time was 43.0m?

Answer 1

Acceleration,
`a=11.62\ m/s^2` (west)

Step-by-step explanation:

Initial velocity, u = 15 m/s

Final velocity, v = 35 m/s

Displacement, d = 43 m

We need to find the magnitude of acceleration. Here, both initial and final velocities are in same direction i.e. west, so we have to take both velocities positive. It can be calculated using third equation of motion as :

`v^2-u^2=2ad`

`a=(v^2-u^2)/(2d)`

`a=((35\ m/s)^2-(15\ m/s)^2)/(2* 43\ m)`

`a=11.62\ m/s^2`

So, the acceleration of the object is
`11.62\ m/s^2` (west) . Hence, this is the required solution.