Question

The velocity vs time graph of an object is approximated by a triangle which starts at v=0 at t=0, rises to a maximum of v=6 m/s at t=6 sec, then returns to zero at t=10 sec. How far did the object travel?

Answer 1

Would I have to use one of the kinematic equations?

Answer 2

You can use the kinematic equations. For the first part,
\[x_1=\frac{1}{2}a_1t_1^2\Rightarrow x_1=18\ m\] And for the second,
\[x_2=v_0t_2+\frac{1}{2}a_2t_2^2=12\ m\]Then, total displacement is 30 m. For the first part you can obtain the value of a
\[a=\Delta v/\Delta t=6/6=1\ m/s^2\] The same for the second,
\[a=\Delta v/\Delta t=-6/4=-3/2\ m/s^2\] But with
\[v_0=6\ m/s\].

Answer 3

Thank you!