# A particle is moving along the x-axis. Its position at time t>0 is e^(2-t). What is its acceleration when t=2?

### 1 Answer | Add Yours

the position in the x direction is given by,

`x = e^((2-t)) --- (1)`

` ` now the acceleration in the x direction is given by, ` ``(d^2x)/dt^2`

lets differenciate the equation (1) to obtain an expression for the acceleration

`x = e^((2-t))`

`dx/dt = -e^((2-t))`

`(d^2x)/dt^2 = e^((2-t))`

` `

` ` when t = 2,` `

`(d^2x)/dt^2 = e^((2-t)) = e^((2-2)) = e^0 = 1`

` <br> `

**Hence the acceleration when t = 2 is 1**

`

`