There is also m at the bottom...sorry about that couldn’t fit it on with the res

Question

There is also m at the bottom...sorry about that couldn’t fit it on with the rest. Thanks

The position of a particle moving along the x axis depends on the time according to the equation x = ct2 bt3, where x is in meters and t in seconds. Units of constant C are m/s^2. Units of constant B are m/s^3.

You have 1 day remaining in your WebAssign Grace Period. Get Access Today. For the following, let the numerical values of c and b be 4.0 and 1.6, respectively. (For vector quantities, i (c) At what time does the particle reach its maximum positive x position? d) From t 0.0 s to t 4.0 s, what distance does the partide move? (e) From t-0.0 s to t 4.0 s, what is its displacement? (1 Find its velocity at t-1.0s rid ts veomays Find its velocity at t2.0 s m/s (h) Find its velodty at t 3.0 s. m/s (1) Find its velocity at t 4.0 s. m/s 1) Find its acceleration at t 1.0s. (1) Find its acceleration at t 3.0 s. (m) Find its acceleration at t 4.0 s

Explanation / Answer

x = c t^2 - b t^3

x = 4 t^2 - 1.6 t^3

(C) x will be max at t for which dx/dt = 0

dx/dt = 8 t - 4.8 t^2 = 0

t ( 8 - 4.8t) = 0

t= 0 Or 1.67 s

Ans: 1.67s s

(D) x(0) = 0 , x(1.67) = 3.7

after that it will turn.

x(4) = -38.4

distance = (3.7 - 0 ) + (3.7 - (-38.4))

= 45.8 m

(E) displacement = - 38.4 m

(F) v = dx/dt = 8 t - 4.8t^2

t = 1 , v = 3.2 m/s

(G) t = 2 .

v = -3.2 m/s

(h) t = 3 , v = -19.2 m/s

(i) t = 4, v = - 44.8 m/s

(j) a = dv/dt = 8 - 9.6t

t =1 , a = -1.6 m/s^2

(k) a = -11.2 m/s^2

(l) a = -20.8 m/s^2

(m) a = - 30.4 m/s^2

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