An object moves in a straight line so that its velocity after t seconds is v(t). Determine the distance traveled by the object on the given time interval. v(t)=3radical(t) - t on [0,36]...??

Question

across · Answer

It is a common fact that the integral of the velocity of an object is the distance it travels. Therefore,$$3\int_{0}^{36}\sqrt{t}dt-\int_{0}^{36}tdt.$$Can you integrate that?

anonymous · Answer

no...

anonymous · Answer

displacement is -216 if that helps...

across · Answer

Where did you get that answer from?

anonymous · Answer

someone helped me earlier but didnt help me with distance. and i submitted that answer and it was right..i just dont know how to get the distance

across · Answer

That's what the above integral evaluates to:$$3\int_{0}^{36}\sqrt{t}dt-\int_{0}^{36}tdt=\left [ 2t^{\frac{3}{2}}\right ]_{0}^{36}-\left [ \frac{t^2}{2}\right ]_{0}^{36}=-216$$

anonymous · Answer

ohhh so what about distance then?

across · Answer

Since the distance can't be a negative value, we need to modify the integral as follows:$$\int_{0}^{9}(3\sqrt{t}-t)dt-\int_{9}^{36}(3\sqrt{t}-t)dt=\frac{27}{2}+\frac{459}{2}=243.$$That should be it (if I didn't make a mistake somewhere).

anonymous · Answer

thank you!

across · Answer

Was that it?