Question

(1) An object moves with constant acceleration 3.26 m/s and over a time interval reaches a final velocity of 10.8 m/s. a. If its initial velocity 5.39 m/s, what is its displacement during the time interval? Submit Answer Tries 0/10 b. What is the distance it travels during this interval? Submit Answer Tries 0/10 c. If its initial velocity is -5.39 m/s, what is its displacement during the time interval? Submit Answer Tries 0/10 d. What is the total distance it travels during the interval in part (c)?

Answer 1

as object is moving with constant acceleration, a= 3.26 m/s2 Final velocity, V = 10.8 m/s Initial velocity, wi = 5.39 m/s Positive so velocities same direction. object Both moves along We ком, V 2 - น + 2 as 2as - (V$-v;? vf?- U? sz 2 a 10.82_5.392 2 2 X 3.26 s & 13.43 m Displacement during this intervent is 13.43 (b) Distance will be same also. So, distance travels during this interval = 13.43 mi c) Now, initial velocity, u = -5.39 m/s final relocity, velocity, V = 10.8 m/s acceleration, a 2 3.26 my 32 CS Scanned with: CamScanner

so, Hore are final and intial Welocity sign different. So, object ariston direction. Then it travels in the Positive direction, first in negative moves T -5.29 to so, we have calculate displacement when object along negative direction. 3-5.39 m/s 12 zu2+ras 22.42 os si 29 2 02-(-5.39) 2 X 3.26 Si su 4.45 m Displacement for positive direction. when , uzo, V210,8 m/s $29 22. u2 2.a 2 1018 2X3.26 17,89 m Displacement of the object is a 1 Sas, + 52 2.-4.45 +17.89 a 13.44 m Scanned with CamScanner cs

(d) Total distance travels during this intervala Isil + 152/ = 4.45 of 17 8 9 2.2.34 m cs Scanned with CamScanner