Question

A particle moving along a straight line has an acceleration which varies according to position as shown. If the velocity of the particle at the position x = -4 ft is V-6 ft/sec, determine the velocity v when the acceleration is zero. (Note: there are several instances when acceleration is zero). a, see Sude=adso r -) = area under as curve)

Answer 1

a (fill see?) x+4 5-(-4) 14-(-6) 4+06 - 9 06 20 ki 4+4 = 9X22,74 10 - 4. 20 x = 2.7-4 x(ft → x= -1. 3ft S, when x= -1.3ft acceleration azo Now, fvdva fads - Aree under V s Cars x) At sro - 4 ft V=oftler V 26 ft Judva 26 x 2 (4-1.3) here sec 2 n V-36 16.2 = 52.2 - 1.3ft accelestion aso to Answers Answer when a 2 = 7. 22ft see a velocity

Now when tsag ft accolestion azo kisst find velocity when aa5ft. I v dr= Area under aro x²) here, so = -4 ft de sp=5ft At r=-4 ft Vj = 6 f three f v dr- 1 (6) (4-1.3 + 1 (6.3) (14) Vy 8.1 + 44.1 a 52.2 6 = 104,4 % = 140,4 i. Ty z 11,85 filseal a relucity when aasft 55ast ft accelection as14 ft u du = ha da a Arlanader 5 ft. 4 7ft az 14 ft ₂ = 11,85 All see here

- 14 (7-5) 1 (V, _ (11,858) 28 v² - 140.4 225 = 56 V = 196.4225 144 = 14,015 ft sec velocity when I = 7 ft Now, for 75x s9 ft accelention azo So, velocity will be remains - contant So, Velocity v when acceleration is Zero fon 73729 ft &fazer