The position of a particle is defined by the equation s = 0.12t3 + 0.9t2 - 4t + 6, where s is in metres and t is in seconds. This equation is valid for the time range t=0 to t=5 seconds. Determine the following:(a) the velocity of the particle at t=1 second, v = Answer m/s(b) the acceleration of the particle at t=1 second, a = Answer m/s2
Question
The position of a particle is defined by the equation s = 0.12t3 + 0.9t2 - 4t + 6, where s is in metres and t is in seconds. This equation is valid for the time range t=0 to t=5 seconds. Determine the following:(a) the velocity of the particle at t=1 second, v = Answer m/s(b) the acceleration of the particle at t=1 second, a = Answer m/s2
Solution 1
To solve this problem, we need to use calculus.
(a) The velocity of the particle at any time t is given by the derivative of the position function s(t). So, we need to find the derivative of s = 0.12t^3 + 0.9t^2 - 4t + 6.
The derivative of s with respect to t, denoted as ds/dt or s'(t), is given by:
s'(t) = 0.36t^2 + 1.8t - 4
This is the velocity function. To find the velocity at t = 1 second, we substitute t = 1 into the velocity function:
v = s'(1) = 0.36(1)^2 + 1.8(1) - 4 = 0.36 + 1.8 - 4 = -1.84 m/s
(b) The acceleration of the particle at any time t is given by the derivative of the velocity function v(t). So, we need to find the derivative of v = 0.36t^2 + 1.8t - 4.
The derivative of v with respect to t, denoted as dv/dt or v'(t), is given by:
v'(t) = 0.72t + 1.8
This is the acceleration function. To find the acceleration at t = 1 second, we substitute t = 1 into the acceleration function:
a = v'(1) = 0.72(1) + 1.8 = 2.52 m/s^2
So, the velocity of the particle at t = 1 second is -1.84 m/s and the acceleration of the particle at t = 1 second is 2.52 m/s^2.
Solution 2
To solve this problem, we need to use calculus.
(a) The velocity of the particle at any time t can be found by taking the derivative of the position function with respect to time. This gives us the velocity function:
v(t) = ds/dt = 0.36t^2 + 1.8t - 4
To find the velocity at t=1 second, we substitute t=1 into the velocity function:
v(1) = 0.36(1)^2 + 1.8(1) - 4 = 0.36 + 1.8 - 4 = -1.84 m/s
(b) The acceleration of the particle at any time t can be found by taking the derivative of the velocity function with respect to time. This gives us the acceleration function:
a(t) = dv/dt = 0.72t + 1.8
To find the acceleration at t=1 second, we substitute t=1 into the acceleration function:
a(1) = 0.72(1) + 1.8 = 2.52 m/s^2
Therefore, the velocity of the particle at t=1 second is -1.84 m/s and the acceleration of the particle at t=1 second is 2.52 m/s^2.
Similar Questions
The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2, v(0) = −15, 0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = t2+2t−15 m/s(b) Find the distance traveled during the given time interval.
A particle moves according to the equation; x = 10t2, where x is in meters and t is in seconds. Find the velocity for the time interval from 2.0 s to 2.1 s. (a) 0.1 m/s (b) 42 m/s (c) 44.1 m/s (d) 40 m/s (e) 2.0 m/s 11
The position of a particle related to time is given by x = ( 5 t 2 − 4 t + 5 ) m . The magnitude of velocity of the particle at t = 2 s will be :
The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8, 0 ≤ t ≤ 5(a) Find the displacement. m(b) Find the distance traveled by the particle during the given time interval. m
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.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.