Question

When an object is dropped straight down, the distance in feet) that it travels in t seconds is given by s(t) = - 1612. Find the velocity at each of the following times. a. After 2 seconds. b. After 6 seconds. c. After 10 seconds. d. Find the acceleration. a. The velocity after 2 seconds is ft/sec.

Answer 1

Plotting: y = x° – 15x“ + 75x – 1 2001 -10 10 -200

Plotting: y = X - 2 X+ 3 5 -20 20 -5

Ans s(t) = - 16t? an object is the rate at which we know that the velocity of it changes its position..so,. (t) = s'lt) - 32(t). velocity after a seconds ; N(2) = -32x2 = -64 ft/s. (6 : after 6 seconds. V (6) = -32X6 -192 ft/s after 10 seconds j v (10) = -32% Iu - 320 *t/s We know that the acceleration is the rate of change of velocity, sos act) = v'CH) = -32 ft/s? d 7-5 y=w x315x²+750 + 750 -1. 3) y'(X) = 322- 3071 +75 (x-5) (71-5) is the only, = coitical poruß are y'W=0 critical penut. **5 y'amo is positive en el 5 y'an? ſs positive therebere. is a saddle porut. not a maxima ar minima. 3_15a²+754-1 has no. maxima . 1 Thu at at 21-5 - yCa)= 71-2 5 3 f(n)= from 2 = 1+3 (+3) & to find Critical point take f(2)=0 5 520 not possible. (743)? 2-0 therefore Thus no canitical points. fras has +69) has relative maxima or minima. no