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