Suppose that the position of a particle is given by s(t)=4t3+5t+9. (a) Find the velocity at time t. v(t)=?mc (b) Find the velocity at time t=−3sec⁡onds. _____ ms (c) Find the acceleration at time t. a(t)=?ms2 (d) Find the acceleration at time t=3sec⁡onds. _____ ms2

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Thomas Lynn · Accepted Answer

Step 1 Given  Position of a particle is given by s(t)=4t3+5t+9  Step 2 (a) Velocity at time t  Given: Position s(t)=4t3+5t+9  Velocity v(t) is given by: v(t)=d[s(t)]dt  v(t)=d[s(t)]dt  v(t)=d(4t3+5t+9)dt  4(3)t3−1+5(1)+0  =12t2+5  Therefore, velocity at time t is: v(t)=(12t2+5)ms

Joseph Lewis · Accepted Answer

Step 3 (b) Velocity at time t=3 seconds  Velocity at time t is: v(t)=12t2+5  Therefore, velocity at time t=3 seconds is given by  v(t)=12t2+5  v(3)=12(3)2+5  =12(9)+5  =108+5  =113  Therefore, velocity at time t=3 seconds is:  v=113ms

karton · Accepted Answer

Step 4 (c) Acceleration at time t Velocity at time t is: v(t)=12t2+5 Acceleration a(t) is given by: a(t)=d[v(t)]dt a(t)=d[v(t)]dt a(t)=d(12t2+5)dt =12(2)t2−1+0 =24t Therefore, acceleration at time t is: a(t)=24tms2 Step 5 (d) Acceleration at time t=3 seconds Acceleration at time t is: a(t)=24t Therefore, acceleration at time t=3 seconds is given by a(t)=24t a(3)=24(3) =72 Therefore, acceleration at time t=3 seconds is: a=72ms2

xleb123 · Accepted Answer

(a) To find the velocity at time t, we need to differentiate the position function s(t) with respect to time. Using the power rule for differentiation, we have:v(t)=ds(t)dt=ddt(4t3+5t+9)=12t2+5 ms(b) To find the velocity at time t = -3 seconds, we substitute t = -3 into the velocity function:v(−3)=12(−3)2+5=108+5=113 ms(c) To find the acceleration at time t, we need to differentiate the velocity function v(t) with respect to time:a(t)=dv(t)dt=ddt(12t2+5)=24t ms2(d) To find the acceleration at time t = 3 seconds, we substitute t = 3 into the acceleration function:a(3)=24(3)=72 ms2

Andre BalkonE · Accepted Answer

Step 1:(a) To find the velocity at time t, we need to differentiate the position function s(t) with respect to time. Using the power rule of differentiation, we obtain:v(t)=dsdt=ddt(4t3+5t+9)Taking the derivative of each term separately, we have:v(t)=ddt(4t3)+ddt(5t)+ddt(9)Applying the power rule, we get:v(t)=12t2+5Therefore, the velocity at time t is given by v(t)=12t2+5ms.Step 2:(b) To find the velocity at time t = -3 seconds, we substitute t = -3 into the velocity function we derived in part (a). Plugging in t = -3, we have:v(−3)=12(−3)2+5Simplifying the expression:v(−3)=12(9)+5v(−3)=108+5v(−3)=113msTherefore, the velocity at time t = -3 seconds is v(−3)=113ms.Step 3:(c) To find the acceleration at time t, we need to differentiate the velocity function v(t) with respect to time. Taking the derivative of v(t), we get:a(t)=dvdt=ddt(12t2+5)Differentiating each term separately, we obtain:a(t)=ddt(12t2)+ddt(5)Using the power rule, we have:a(t)=24tThus, the acceleration at time t is given by a(t)=24tms2.Step 4:(d) To find the acceleration at time t = 3 seconds, we substitute t = 3 into the acceleration function we derived in part (c). Substituting t = 3, we have:a(3)=24(3)a(3)=72ms2Therefore, the acceleration at time t = 3 seconds is a(3)=72ms2.

Suppose that the position of a particle is given by s(t)

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