If the jerk is 2, then acceleration is 2t, velocity is $t^2$ and distance is $t^{3}/3$? [closed] Are my graphs and equations right? I just began learning about these..

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

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

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.