Question

5) The acceleration function (in m/s2) and the initial velocity are given for a particle moving along a line. (3 points) a(t) = 5t + 2, v(0) = 6, Osts 4 a) Find the velocity at time t. b) Find the distance traveled during the given time interval.

Answer 1

5) 5t+2 (a) Given that act) dv (t)_ 56+2 at Integrating we have Vlt) = Şt + at tc Using the condition, v10) = 6 6 - 0 +o+c => c = 6 velocity at time t. 19(+) t 2t + 6 , is (6) dt dB (H) pttat +6 9. Sase) = fazer at 46) 4 2 févat +25*'* 40 fa t=0 t=0 dt st = t=0 4 4 +6 [t] 2[ 5. t [ ] is the required distance. 280 3 §. 4 + 4 +6.4 = 6

X 2 at 2) 2 e x² 2 2 F(x) = S et F(a) = .: F (2) adso, F(2) = f*at 20 -- Equation of tangent for 2 = 2; 4 ze e un y - F(2) - F'(2) 2-2 > Y-O 4 => y = e'x - 264 X-2