Kinematic Calculus

Classical Mechanics Level 2

A car, starting at rest at t = 0 t = 0 , moves at a speed of e t / 2 m / s e^{t/2}\ \rm m/s . How far has the car moved in 10 s 10\ \rm s ? Give your answer to the nearest m \rm m .

Notation: e ≈ 2.718 e \approx 2.718 denotes the Euler's number .


The answer is 295.

Kano Boom by Kano Boom , 16, Australia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The distance of the car after 10 s 10 \text{ s} would be

∫ 0 10 e t 2 d t = 2 e t 2 ∣ 0 10 = 2 ( e 5 − 1 ) ≈ 294.8263... ⟹ Answer = 295 \displaystyle \int _0^{10}\:e^{\frac{t}{2}}dt = \left. 2e^{ \frac{t}{2}} \right|_0^{10} = 2(e^{5} -1) \approx 294.8263... \implies \text{Answer} = \blue{\boxed{295}}

5 Helpful 1 Interesting 2 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...