Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. a(t)=2i+2tk,v(0)=3i−j,r(0)=j+k

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Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.  a(t)=2i+2tk,v(0)=3i−j,r(0)=j+k

ensojadasH · Accepted Answer

v(t)=∫a(t)dt  Given that  a(t)=2i+2tk  Therefore  v(t)=∫2i+2tkdt  v(t)=2ti+t2k+c  Where C is constant vector.  Given that v(0)=3i−j. Therefore 3i−j=C  Substitute value of C, to get  v(t)=2ti+t2k+3i−j  v(t)=(2t+3)i−j+t2k  r(t)=(t2+3t)i+(1−t)j+(t33+1)k  The position vector and the velocity vector for the given acceleration vector is r(t)=(t2+3t)i+(1−t)j+(t33+1)k and v(t)=(2t+3)i−j+t2k respectively.

Jazz Frenia · Accepted Answer

Given:a(t)=2𝐢+2t𝐤v(0)=3𝐢−𝐣r(0)=𝐣+𝐤First, let&#039;s find the velocity vector 𝐯(t).Integrating the acceleration, we have:∫a(t)dt=∫(2𝐢+2t𝐤)dt𝐯(t)=∫2𝐢dt+∫2t𝐤dt𝐯(t)=2t𝐢+t2𝐤+C1Next, we&#039;ll find the constant vector C1 using the initial velocity condition 𝐯(0)=3𝐢−𝐣.Substituting t=0 into the velocity equation, we get:𝐯(0)=2(0)𝐢+(0)2𝐤+C1C1=3𝐢−𝐣Therefore, the velocity vector is:𝐯(t)=2t𝐢+t2𝐤+(3𝐢−𝐣)Now, let&#039;s find the position vector 𝐫(t).Integrating the velocity, we have:∫𝐯(t)dt=∫(2t𝐢+t2𝐤+(3𝐢−𝐣))dt𝐫(t)=∫2t𝐢dt+∫t2𝐤dt+∫(3𝐢−𝐣)dt𝐫(t)=t2𝐢+13t3𝐤+(3t𝐢−t𝐣)+C2Next, we&#039;ll find the constant vector C2 using the initial position condition 𝐫(0)=𝐣+𝐤.Substituting t=0 into the position equation, we get:𝐫(0)=(0)2𝐢+13(0)3𝐤+(3(0)𝐢−(0)𝐣)+C2C2=𝐣+𝐤Therefore, the position vector is:𝐫(t)=t2𝐢+13t3𝐤+(3t𝐢−t𝐣)+(𝐣+𝐤)In short form, the velocity vector 𝐯(t) is 2t𝐢+t2𝐤+(3𝐢−𝐣) and the position vector 𝐫(t) is t2𝐢+13t3𝐤+(3t𝐢−t𝐣)+(𝐣+𝐤).

Andre BalkonE · Accepted Answer

Step 1:First, let&#039;s integrate the acceleration vector to find the velocity vector. Integrating a→(t) with respect to time, we have:v→(t)=∫a→(t)dtIntegrating each component separately, we get:vx(t)=∫2dt=2t+C1vy(t)=∫0dt=C2vz(t)=∫2tdt=t2+C3Here, C1, C2, and C3 are constants of integration. Since we know the initial velocity v→(0)=3𝐢−𝐣, we can substitute t=0 into the expressions above:vx(0)=2(0)+C1=3⟹C1=3vy(0)=C2=−1vz(0)=(0)2+C3=0⟹C3=0Therefore, the velocity vector v→(t) is:v→(t)=(2t+3)𝐢−𝐣+t2𝐤Step 2:Next, let&#039;s integrate the velocity vector to find the position vector. Integrating v→(t) with respect to time, we have:r→(t)=∫v→(t)dtIntegrating each component separately, we get:rx(t)=∫(2t+3)dt=t2+3t+C4ry(t)=∫(−1)dt=−t+C5rz(t)=∫t2dt=13t3+C6Here, C4, C5, and C6 are constants of integration. Using the initial position r→(0)=𝐣+𝐤, we substitute t=0:rx(0)=(0)2+3(0)+C4=0⟹C4=0ry(0)=−(0)+C5=1⟹C5=1rz(0)=13(0)3+C6=1⟹C6=1Therefore, the position vector r→(t) is:r→(t)=(t2+3t)𝐢+(−t+1)𝐣+(13t3+1)𝐤Thus, we have found the velocity vector v→(t) and the position vector r→(t) of the particle in terms of time t.

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