The velocity of a particle is given by the equationv = 9t2 i + t3 j + (0.75t + 3) k m/swhere t is in seconds. Determine the following:(a) the magnitude of the particle’s acceleration when t=1.75 seconds, a = Answer m/s2.(b) the magnitude of the particle's displacement when t=1.75 seconds (r=0 when t=0), r = Answer metres.

To solve this problem, we first need to understand that the acceleration of the particle is the derivative of the velocity with respect to time, and the displacement of the particle is the integral of the velocity with respect to time.

(a) To find the acceleration, we need to take the derivative of the velocity function. The derivative of 9t^2 with respect to t is 18t, the derivative of t^3 is 3t^2, and the derivative of 0.75t + 3 is 0.75. So the acceleration function is a = 18t i + 3t^2 j + 0.75 k m/s^2.

To find the magnitude of the acceleration at t = 1.75 seconds, we substitute t = 1.75 into the acceleration function to get a = 18(1.75) i + 3(1.75)^2 j + 0.75 k = 31.5 i + 9.1875 j + 0.75 k m/s^2.

The magnitude of a vector is given by the square root of the sum of the squares of its components. So the magnitude of the acceleration is sqrt((31.5)^2 + (9.1875)^2 + (0.75)^2) = sqrt(992.109375) = 31.5 m/s^2.

(b) To find the displacement, we need to integrate the velocity function with respect to time from 0 to 1.75. The integral of 9t^2 dt from 0 to 1.75 is (3t^3)|_0^1.75 = 3(1.75)^3 - 3(0)^3 = 16.40625. The integral of t^3 dt from 0 to 1.75 is (0.25t^4)|_0^1.75 = 0.25(1.75)^4 - 0.25(0)^4 = 2.44140625. The integral of (0.75t + 3) dt from 0 to 1.75 is (0.375t^2 + 3t)|_0^1.75 = 0.375(1.75)^2 + 3(1.75) - (0.375(0)^2 + 3(0)) = 4.21875 + 5.25 = 9.46875.

So the displacement function is r = 16.40625 i + 2.44140625 j + 9.46875 k metres.

The magnitude of the displacement is sqrt((16.40625)^2 + (2.44140625)^2 + (9.46875)^2) = sqrt(370.2548828125) = 19.24 metres.