A particle initially at rest moves in a straight line with an acceleration of (10t – 4t\(^2\))m/s\(^2\)
Find the:
a. velocity of the particle after t seconds;
ii. average acceleration of the particle during the 4th second.
b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take g = 10m/s\(^2\)]
ii. with an acceleration of 3m/s\(^2\). [Take g = 10m/s\(^2\)]
Explanation
a. a = (10t - 4t\(^2\))m/s\(^2\), t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t\(^2\))t
v = (10t\(^2\) - 4t\(^3\))m/s\(^2\)
ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3)\(^2\)
Δa4 = 10(4-1) - 4(4-3)\(^2\)
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s\(^2\)
Δa4 = 6ms-2
b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N
ii. when the lift moves upward with an acceleration of 3m/s\(^2\)
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N