The distance covered by a moving particle through point O is given by the equation, s = t^3- 15t^2+ 63t -10.
Find:
(a) distance covered when t = 2
(b) the distance covered during the 3rd second;
(c) the time when the particle is momentarily at rest;
(d) the acceleration when t = 5
b. distance during third second: s(4)-s(3)=4^3-3^3-15(4^2-3^2)+63(4-2)
c. when at rest ds/dt=0 or 3t^2-30t+63=0
solve the quadratic, solve for time t.
d. s"=6t-30 at t=5, zero.
To find the distance covered by a moving particle at a specific time or during a specific interval, we need to substitute the value of time, t, into the given equation, s = t^3 - 15t^2 + 63t - 10.
(a) Distance covered when t = 2:
Substitute t = 2 into the equation: s = (2)^3 - 15(2)^2 + 63(2) - 10.
Calculating the values: s = 8 - 60 + 126 - 10.
Therefore, the distance covered when t = 2 is s = 64 units.
(b) Distance covered during the 3rd second:
To find the distance covered during a specific time interval, we need to calculate the value of s at the end of the interval minus the value of s at the start of the interval. In this case, the interval is from t = 2s to t = 3s, so we need to calculate s at t = 3s and t = 2s.
For t = 3, substitute t = 3 into the equation: s = (3)^3 - 15(3)^2 + 63(3) - 10.
Calculating the values: s = 27 - 135 + 189 - 10 = 71.
For t = 2, substitute t = 2 into the equation: s = (2)^3 - 15(2)^2 + 63(2) - 10.
Calculating the values: s = 8 - 60 + 126 - 10 = 64.
Therefore, the distance covered during the 3rd second is (s at t = 3) - (s at t = 2) = 71 - 64 = 7 units.
(c) Time when the particle is momentarily at rest:
The particle is momentarily at rest when its velocity is zero. To find the time(s) when the particle is momentarily at rest, we need to find the roots of the equation for velocity. Velocity is the derivative of distance with respect to time.
Differentiating the given equation, we get v = ds/dt = 3t^2 - 30t + 63.
To find the roots of the equation, set v = 0:
0 = 3t^2 - 30t + 63.
We can factorize the quadratic equation: 0 = 3(t - 3)(t - 7).
Setting each factor equal to zero, we get t = 3 and t = 7.
Therefore, the particle is momentarily at rest when t = 3s and t = 7s.
(d) Acceleration when t = 5:
Acceleration is the derivative of velocity with respect to time. Differentiating the velocity equation, we get a = dv/dt = 6t - 30.
To find the acceleration when t = 5, substitute t = 5 into the equation: a = 6(5) - 30.
Calculating the values: a = 30 - 30 = 0.
Therefore, the acceleration when t = 5 is a = 0 units/s^2.