A particle moves along the x-axis according to the equation x(t)= 2.9- 4.1t^2 m. what are the velocity (in m/s) and acceleration

Question

irina1246 [14]

3 years ago

13

A particle moves along the x-axis according to the equation x(t)= 2.9- 4.1t^2 m. what are the velocity (in m/s) and acceleration

(in m/s^2) at t=3.0 s and t=6.0 s

Physics

1 answer:

kherson [118] · Answer 1 · 2022-01-05T03:42:18+03:00

From the equation you can tell that your particle is moving with a costant acceleration. The generic movement equation is $x= x_0 +v_0 t + \frac 1 2 a t^2$ , and velocity is given by $v= v_0 + at$ with $x_0$ and $v_0$ being the position and velocity at $t=0$ s.

By comparing the two acceleration is costant, and it's the same at t=0, t=3.0, or t=3000s, and is $\frac 1 2 a = 4.1$ . Velocity comes from the second formula, you found the acceleration in the previous step, and you just have to plug 0 for $v_0$ , the value of a you found, and 6 s for t.