An object has a velocity v 0 ( t ) = (3.00 m/s) i + (1.00 m/s) j at time t = 0.0

Question

An object has a velocity v0(t) = (3.00 m/s) i  + (1.00 m/s) j  at time t = 0.00 s. The acceleration of the object is a(t) = (2.00 m/s3)t i  + (1.00 m/s4)t2 j . If the object is located at r0(t) = (1.00 m) i  + (-1.00 m) j at time t = 0.00 s., what is the position of the object at time t = 1.00 s?

choices :

(5.32 m) i  + (1.34 m) j

(4.33 m) i  + (0.0833 m) j

(2.72 m) i  + (6.14 m) j

(0.333 m) i  + (0.0833 m) j

(2.00 m) i  + (1.00 m) j

choices :

(5.32 m) i  + (1.34 m) j

(4.33 m) i  + (0.0833 m) j

(2.72 m) i  + (6.14 m) j

(0.333 m) i  + (0.0833 m) j

(2.00 m) i  + (1.00 m) j

Explanation / Answer

Vo (t) = 3 i^ + 1 j^

a(t) = 2 t i^ + 1 t2 j^

dV(t)/dt = 2 t i^ + 1 t2 j^

V(t) = t2 i^ + (t3/3) j^ + C

at t = 0

V(0) = C

3 i^ + 1 j^ = C

so V(t) = t2 i^ + (t3/3) j^ + 3 i^ + 1 j^

dx(t)/dt = t2 i^ + (t3/3) j^ + 3 i^ + 1 j^

X(t) = (t3/3) i^ + (t4/12) j^ + 3t i^ + t j^ + D

at t = 0

X(0) = D

D = 1 i^ - 1 j^

X(t) = (t3/3) i^ + (t4/12) j^ + 3t i^ + t j^ + 1 i^ - 1 j^

at t = 1

X(1) = (13/3) i^ + (14/12) j^ + 3 i^ + j^ + i^ - j^

X(1) = 4.33 i^ + 0.0833 j^

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