Question

The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt4, where x is in meters and t in seconds. Let c and b have numerical values 2.6 m/s3 and 1.7 m/s4, respectively. From t = 0.0 s to t = 1.7 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.

Answer 1

\(Displacement(t) = x(t_{Stop}) - x(t_{Start})\)

\(Velocity(t) = x'(t) = \dfrac{dx}{dt}\)

Let's see what you get.

Answer 2

displacement in a is -1.4?

Answer 3

@tkhunny and the velocity in b is 1?

Answer 4

in order to get the acceleration, it's the second derivative and then put the values in the equation is that right?

Answer 5

Hint:
\[\begin{gathered}
x\left( 0 \right) = c \cdot 0 - b \cdot 0 = 0 \hfill \\
x\left( {1.7} \right) = c \cdot {\left( {1.7} \right)^3} - b \cdot {\left( {1.7} \right)^4} = 2.6 \cdot {\left( {1.7} \right)^3} - 1.7 \cdot {\left( {1.7} \right)^4} = ...? \hfill \\
\end{gathered} \]

Answer 6

Did you find v(t) = x'(t)? That would be the place to start.

Answer 7

yeah v(t) = 3ct^2 - 4bt^3 ?

Answer 8

so acceleration is a(t) = x"(t) = 6ct^2 - 12bt^2 ?

Answer 9

ok!

Answer 10

@Michele_Laino so my answer is right on letter a ? it should be -1.4? :)

Answer 11

or -1.42

Answer 12

yes!

Answer 13

-1.42

Answer 14

thank you guys :D

Answer 15

thank you! :)