An object is moving along a line with velocity v (t )equals A t squared space plus space B meters per second at time t seconds. Its initial position, at time t equals 0, is s (0 )equals C meters. If A equals 15, B equals 8 and C equals space 3, then Give the object's position function s (t ). Give the distance traveled by the object between t equals 1 and t equals 2 seconds and record your answer below.

- S(t) = 5t³+8t+3

- 19.7m approximately

Explanation:

Given the velocity if an object to be

v(t) = At² + B where t is the time in seconds

Velocity is the change in displacement of a body with respect to time.

V(t) = dS(t)/dt

Making S(t) the subject of the formula

dS(t) = v(t)dt

Integrating both sides

∫dS(t) =∫v(t)dt

S(t) = ∫(At²+B)dt

S(t) = At³/3+Bt + C... 1

In its initial position s(0) = 0

t = 0, s = 0

S(0) = A(0)³/3+B(0)+C

S(0)= C

a) The object's position function s(t) if A = 15, B = 8 and C = 3 can be gotten by substituting this value into eqn 1

S(t) = 15t³/3+8t+3

S(t) = 5t³+8t+3

b) For the distance traveled by the object between t equals 1 and t equals 2 seconds

When t = 1s

S(1) = 5(1)³/3+8(1)+3

S(1) = 5/3+8+3

S(1) = 5+24+9/3

S(1) = 38/3 m

When t = 2secs

S(2) = 5(2)³/3+8(2)+3

S(2) = 40/3+16+3

S(2) = (40+48+9)/3

S(2) = 97/3

Distance travelled between this times will be 97/3-38/3

= 59/3

= 19.7m approximately

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y