-2
$\begingroup$

enter image description here

Are my graphs and equations right? I just began learning about these..

$\endgroup$

closed as off-topic by Aaron Stevens, WillO, rob Jan 28 at 14:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, WillO, rob
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ You need constants of integration: $\int 2\,dt = 2t + c$ and so on. $\endgroup$ – tfb Jan 28 at 14:12
  • $\begingroup$ Or you need the initial and final value of $t$. $\int_{t_i}^{t_f} 2 = 2 (t_f-t_i)$ $\endgroup$ – harshit54 Jan 28 at 14:14
1
$\begingroup$

Not in general. You're forgetting all of the constants of integration.

Suppose you start with a constant jerk $j$. Then the acceleration $a$ as a function of time is

$$a(t)=\int j\;dt=jt+a_0$$

where $a_0$ is the initial acceleration. This means that the velocity $v$ as a function of time is

$$v(t)=\int a(t)\;dt=\int (jt+a_0)\;dt=\frac{1}{2}jt^2+a_0t+v_0$$

where $v_0$ is the initial velocity. Likewise, the position $x$ as a function of time will be

$$x(t)=\int v(t)\;dt=\int\left(\frac{1}{2}jt^2+a_0t+v_0\right)\;dt=\frac{1}{6}jt^3+\frac{1}{2}a_0t^2+v_0t+x_0$$

where $x_0$ is the initial position. So, for motion with constant jerk to be well-defined, you need to specify an initial position, velocity, and acceleration.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.