Answer:
a) ax = (-0.036a)t m/s^3; ay = 0.55 m/s^2
b) |v| = 7.469 m/s; Θ = 59°
c) |a| = 0.62 m/s^2; Θ = 298°
Explanation:
a)
the instantaneous acceleration equation will be equal to:
a = dv/dt
ax(t) = (d*(5-0.018*t^2))/dt = -0.036*t
ay(t) = (d*(2+0.55*t))/dt = 0.55
a = (-0.036*t)i + 0.55j
b)
at a time of 8 seconds, the speed of the vector will be equal to:
v = (5 - 0.018*8^2)i + (2 + 0.55*8)j = 3.85i + 6.4j
the magnitude of the vector will be equal to:
|v| = (vx^2 + vy^2)^1/2 = (3.85^2 + 6.4^2)^1/2 = 7.469 m/s
the direction of the vector is equal to:
Θ = tan-1(vy/vx) = tan-1(6.4/3.85) = 59°
c)
the vector acceleration at a time of 8 seconds will be equal to:
a = (-0.036*8)i + 0.55j = -0.288i + 0.55j
the magnitude will be equal to:
|a| = (ax^2 + ay^2)^1/2 = ((-0.288^2) + 0.55^2)^1/2 = 0.62 m/s^2
the direction:
Θ = tan-1(ay/ax) = tan-1(0.55/-0.288) = 298°
