# Thread: who can solve this question?!!

1. ## who can solve this question?!!

The trajectory of a moving particle is given by, y = sin(x) . exp(.2x) .
Then, the acceleration of the particle is:
• cos(x)+4exp^(-2x)
• -cos(x)+4exp^(-2x)
• -sin(x)-2exp^(-2x)
• sin(x)-2exp^(-2x)
• NON of mentioned

2. Originally Posted by soso love
The trajectory of a moving particle is given by, y = sin(x) . exp(.2x) .

Then, the acceleration of the particle is:

• cos(x)+4exp^(-2x)
• -cos(x)+4exp^(-2x)
• -sin(x)-2exp^(-2x)
• sin(x)-2exp^(-2x)
• NON of mentioned

$\displaystyle v(t)=2sin(x)e^{2x}+cos(x)e^{2x}\Rightarrow{x'(t)}$...and then $\displaystyle a(t)\Rightarrow{v'(t)}\Rightarrow{4sin(x)e^{2x}+2c os(x)e^{2x}+2cos(x)e^{2x}-sin(x)e^{2x}}$

3. ## non

Is it $\displaystyle sin(x)exp(-2x)$ or $\displaystyle sin(x)-exp(-2x)$ or what ? What do the . refer to ?