The acceleration of a particle moving only on a horizontal xy plane is given by a- -3t- -4t- where a- is in meters per second squared and t is in seconds- At t-0- the position vector r- -20-0m- -40-0m- - locates the particle- which then has the velocity vector v- -5-00m-s- -2-00m-s- - At t-4-00 s- what are a its position vector in unit vector notation and b the angle between its direction of travel and the positive direction of the x axis-

Using a⃗ =3tî+̂ 4tĵ,̂ we have (in m/s) #160; v⃗ (t)=v⃗ ⃗_⃗ ⃗0⃗ +∫ _ 0 ^ 1 a⃗ #160; dt=(5.00î +2.00ĵ )+∫ _ 0 ^ 1 (3tî +4tĵ ) dt= (5.00+3 t ^ 2 ...

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Standard XII

Physics

Acceleration in 2D

The accelerat...

Question

The acceleration of a particle moving only on a horizontal $x y$ plane is given by $\to a = 3 t^i + 4 t^j,$ where $\to a$ is in meters per second squared and $t$ is in seconds. At $t = 0$ , the position vector $\to r = (20.0 m)^i + (40.0 m)^j .$ locates the particle, which then has the velocity vector $\to v = (5.00 m / s)^i + (2.00 m / s)^j .$ At $t = 4.00 s,$ what are (a) its position vector in unit vector notation and (b) the angle between its direction of travel and the positive direction of the $x$ axis?

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Solution

Using $\to a = 3 t^i + 4 t^j,$ we have (in m/s)
$\to v (t) =_{0} + \int_{0}^{1} \to a d t = (5.00^i + 2.00^j) + \int_{0}^{1} (3 t^i + 4 t^j) d t = (5.00 + 3 t^{2} / 2)^i + (2.00 + 2 t^{2})^j$
Integrating using Eq. then yields (in meters)
$\to r (t) =_{0} + \int_{0}^{1} \to v d t = (20.0^i + 40.0^j) + \int_{0}^{1} [(5.00 + 3 t^{2} / 2)^i + (2.00 + 2 t^{2})^j] d t$
$= (20.0^i + 40.0^j) + (5.00 t + t^{3} / 2)^i + (2.00 t + 2 t^{3} / 3)^j$
$= (20.0 + 5.00 t + t^{3} / 2)^i + (40.0 + 2.00 t + 2 t^{3} / 3)^j$
(a) At $t = 4.00 s$ , we have $\to r (t = 4.00 s) = (72.0 m)^i + (90.7 m)^j .$
(b) $\to v (t = 4.00 s) = (29.0 m / s)^i + (34.0 m / s)^j .$ Thus, the angle between the direction of
travel and $+ x$ , measured counterclockwise, is $θ = {t a n}^{- 1} [(34.0 m / s) / (29.0 m / s)] = {49.5}^{o}$

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