Question

Hi i have an integration question:

A particle moves in a straight line such that after t seconds its velocity is v m/s where v=4-4t-15t^2.
Find the total distance moved by the particle in
(a) The first two seconds.
(b) The third second of motion

THE ANSWER SHOULD BE 41.92M THANKS

Answer 1

should be?

Answer 2

and 101m

Answer 3

for b

Answer 4

integral of v = distance

Answer 5

v(t) = 4-4t-15t^2

example
v(5) = 5 - 4(5) - 15(5)^2

Answer 6

it should be s(t) because s denotes distance really, but this is in terms of velocity

Answer 7

i have integrated to get s=4t-2t^2-5t^3

Answer 8

no

Answer 9

s(t) = int(v(t)) dt = 4-4t-15t^2 <--- this is already a function of displacement (distance) with respect to time
s'(t) = v(t) -4 - 30t
v'(t) = a(t) = 30

Answer 10

ok

Answer 11

try plugging in to your calculator and see if you get the answers you denote

Answer 12

I think there's some missing info. You can't get the displacement equation unless you have some known distance at a certain time.

Integrating \(v(t)\), you have
\[\begin{align*}s(t)&=\int v(t)~dt\\
&=\int (4-4t-15t^2)~dt\\
&=4t-2t^2-5t^3+C
\end{align*}\]
It looks like you assume \(s(0)=0\), then indeed \(C=0\).

However, there is a difference between displacement and distance. \(s(t)\) gives displacement - a net quantity, but \(|s(t)|\) gives distance - a gross total quantity.

To find distance traveled over an interval [a,b], you would instead compute
\[|s(t)|=\int_a^b|v(t)|~dt\]

Answer 13

go back to the formal definitions and equations of velocity and acceleration

Answer 14

\[v=\frac{ \Delta s }{ \Delta t }; a=\frac{ \Delta v }{ \Delta t }\]
when applying to your calculus problem, use d instead of Delta

Answer 15

But distance = rate * time, where the rate is speed, and not velocity.

Compare:
\[\int_0^2v(t)~dt=-40\\
\int_0^2|v(t)|~dt=41.92\]

Answer 16

Simply said:

Distance is not displacement. To find displacement, you do not count the fact that an object is changing directions. But when your're searching for distance, you have to count that fact.

Find out the times where velocity is zero by setting \(v(t) = 0\) and solving the equation. You will get \(\text{time where velocity is zero}= t_0 = \frac{2}{5} \ \text{sec}\).

Wherever you find it to be zero, it means that your object is changing its direction. Integrate this function with proper limits.

For the first two seconds, this is the distance covered:\[\int_{0}^{2/5} v(t)dt + \int_{2/5}^{2}v(t)dt\]The distance covered is NOT\[\int_{0}^{2} v(t)dt\]

Answer 17

if i integrate from 2/5 to 2 should the answer be 41.92

Answer 18

distance after 2 seconds
http://www.wolframalpha.com/input/?i=integrate+4-4x-15x%5E2+dx+from+0+to+.4+%2B+integrate+-%284-4x-15x%5E2%29+dx+from+.4+to+2

distance after 3 seconds
http://www.wolframalpha.com/input/?i=integrate+4-4x-15x%5E2+dx+from+0+to+.4+%2B+integrate+-%284-4x-15x%5E2%29+dx+from+.4+to+3

Answer 19

I'll try to give you a tip.

Therefore, in your question where we are supposed to find displacement/distance from velocity , we need to INTEGRATE.
If you still face difficulty, please tell.