Question

An object moves with constant acceleration 4.70 m/s² and over a time interval reaches a final velocity of 13.6 m/s.

(a) If its original velocity is 6.80 m/s, what is its displacement during the time interval?


(b) What is the distance it travels during this interval?


(c) If its initial velocity is −6.80 m/s, what is its displacement during this interval?


(d) What is the total distance it travels during the interval in part (c)?

Answer 1

Given :- a = 4.7 m/s?, final velocity V=13'6 mis a) initial velocity = 680 mis displacement ? 6.domis 13.6 m/s a = 4.7 m/s² t = 1 S = ? using equation us ut at => t = load to 13-6-618 - 1.4468 4.7 -1.4468 t. 1.456 so s=ut + ļ at? = 6.8% (145) + }(4-7)(1:45) IS. 14. m for straight line motion (displacement) (distancal bo distance - 14.8m - b) c) as initial object Laving opposite velocity to u aceleratin ti 4,5-6.8omis ų 0 si Alinitial) B a:4.7 mis? se (final) U-13.6 m/s Ez 9.4.7 m/s2 - S, -

for AB 1154, -6.80m's 9:491 mis? t> 0-4 , V-0 = 42 tnt, , SES; tj = 0 - (-68) od 4465 Lp tra luss Si = lt + at? = litit. I at? = 668) (145) + }(4.7) (1•45) = -9.86 + 4.943 S, 7-4192 m = AB for B toc , u=42=0 - U=13.6mis a=4.7 m/s², titz t₂ = v. 4₂ t2 = 13-6-0 47 t₂ =2.89 s So Sz = Uzt + ₂ at? = 0 + 4 (4.7)(2-89 Sz=19.65m . So displacement = AC = |BC_LABI {magnitude) = 1521-15,1 = 19.65m 4.92 = 14.73m

or we com also write total displacement as rector sum, with direction (sign) 8 = sits = 1965 – 4892 13= 14.73m) long A to d) total distance travelled = 1 AB) + IBC) - =15, lt 1521 = 4,92'+19.65 distance = 24.57 m