For the following, let the numerical values of c and b be 6.2 and 2.6, respectiv

Question

For the following, let the numerical values of c and b be 6.2 and 2.6, respectively. (For vector quantities, indicate direction with the sign of your answer.)

(c) At what time does the particle reach its maximum positive x position?
s

(d) From t = 0.0 s to t = 4.0 s, what distance does the particle move?
m

(e) From t = 0.0 s to t = 4.0 s, what is its displacement?
m

(f) Find its velocity at t = 1.0 s.
m/s

(g) Find its velocity at t = 2.0 s.
m/s

(h) Find its velocity at t = 3.0 s.
m/s

(i) Find its velocity at t = 4.0 s.
m/s

(j) Find its acceleration at t = 1.0 s.
m/s2

(k) Find its acceleration at t = 2.0 s.
m/s2

(l) Find its acceleration at t = 3.0 s.
m/s2

(m) Find its acceleration at t = 4.0 s.
m/s2

Explanation / Answer

x = ct^2 - bt^3

c = x / t^2 = m/s^2

b = m/s^3

x = 6.2t^2 - 2.6t^3

c) for maximum x will when dx/dt = 0

2(6.2t) - 3(2.6t^2) = 0

12.4t - 7.8t^2 = 0

t = 0 or 1.59 s

so at t = 1.59 s

d) from 0 to 1.59s

x1 = 6.2(1.59^2) - 2.6(1.59^3) = 5.22 m

from 1.59 s to 4 s

x2 = 6.2 ( 4^2 - 1.59^2) - 2.6(4^3 - 1.59^3) = - 72.42 m

distance = 72.42 + 5.22 =77.64 m

e) dispalcement: x = 6.2(4^2) - 2.6(4^3) = - 67.2 m

f) v= dx/dr = 12.4t - 7.8t^2

t = 1s

v = 4.6 m/s

g) at t= 2s

v = - 6.4 m/s

h) t = 3s

v = - 33 m/s

j) a = dv/dt = 12.4 - 2(7.8t) = 12.4 - 15.6t

at t=1

a = - 3.2 m/s^2

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