Physics, 19.09.2019 16:30 Jenniferlam12

Aparticle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second). (a) find the displacement of the particle during 1 ≤ t ≤ 10. (b) find the distance traveled during this time period.

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Answer from: misslux

(a) The displacement will be "103.5".(b) The distance travelled will be "48.166667".

(a)

According to the question,

$1 \leq t \leq 10$

By applying integration, we get

= $\int\limits^{-10}_1 {[t^2-t-20]} \, dx$

= $[\frac{t^3}{3} -\frac{t^2}{2} -20t]_1^{10}$

= $\frac{1}{3}[10^3-1^3]-\frac{1}{2} [10^2-1^2] -20[10-1]$

= $\frac{1}{3}[1000-1]-\frac{1}{2} [99]-20[9]$

= 333-49.5-180

= 103.5

(b)

According to the question,

t^2-t-20 = (t-5)(t+4)

Now,

On interval [1, 5], v_t < 0

On interval [5, 10], v_t 0

→ $\int\limits^{10}_1 {v(t)} \, dt = \int\limits^5_1 {[-v(t)} \, dt]+ \int\limits^{10}_5 v(t) \, dt$

$= \int\limits^5_1 {[-t^2+t+20]} \, dt + \int\limits^{10}_5 {[t^2+t+20]} \, dt$

$=[\frac{-t^3}{3} ]_1^5+[\frac{t^2}{2} ]_1^5+20[t]_1^5+[\frac{t^3}{3}-\frac{t^2}{2} -20t ]_5^{10}$

$= (\frac{1}{3}\times 125-\frac{1}{2}\times 25 )+(\frac{1}{3} -\frac{1}{2} -20)+(\frac{1}{3}\times 1000-\frac{1}{2}\times 100-20\times 10)- (\frac{1}{3}\times 125-\frac{1}{2}\times 25-20\times 5 )$

$= \frac{289}{6}$

= 48.166667

Thus the responses above are correct.

Learn more about displacement here:

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Answer from: ShahinF7536

Displacement is 47,166667 meters, Distance traveled is 78.5 meters.

Explanation:

Let's remberi that $v(t) = \frac d{dt}s(t)$ . Distance traveled will be $\int\limits^{10}_1 v(t) \, dt$ , while displacement will be $\int\limits^{10}_1 |v(t)| \, dt$ (note the absolute value is needed because part of the movement MIGHT be backwards, which will add to the distance traveled, ie the number of steps, but does not contribute to displacement - think of a guy running in cirles).

Distance: $\int\limits^{10}_1 {t^2-t-20} dt =(\frac13 t^3 - \frac12 t^2 -20t )\limits^{10}_1 = (\frac 13 *1000 - \frac12*100-20*10) - (\frac13 -\frac 12 -20) = \frac{2000-300-1200-2+3+60}6 = \frac{561}6 =78.5$

Displacement: absolute value makes things a bit more annoying: v(t) is negative between t=1 and t=5 - the particle is moving backwards. The integral becomes:

$\int\limits^{10}_1 |v(t)| \, dt = -\int\limits^5_1t^2-t-20\,dt +\int\limits^{10}_5t^2-t-20\, dt =\\-(\frac13 t^3 -\frac12 t^2-20t)\limits^5_1 +(\frac13 t^3 -\frac12 t^2-20t)\limits^{10}_5 =\\ -(\frac13*125-\frac1/2*25-20*5)+(\frac13-\frac12-20)+(\frac13*1000-\frac12*100-20*10)-(\frac13*125-\frac1/2*25-20*5) = 289/6 = 48,166667$

Puff, pant, that should be it, please double check calculations, just in case.

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