Hi, please, can explain step by step. A car starts from rest and accelerates in
ID: 1764219 • Letter: H
Question
Hi, please, can explain step by step. A car starts from rest and accelerates in a straight line atan acceleration given by a = bt0.50 , where b = 4.5 m/s2.5 a. What is its acceleration and velocity at times t =1.0 s ,2.0 s, and 5.0 s ? b. How far does it travel in the time interval t = 0 to t =3.0 s? Hi, please, can explain step by step. A car starts from rest and accelerates in a straight line atan acceleration given by a = bt0.50 , where b = 4.5 m/s2.5 a. What is its acceleration and velocity at times t =1.0 s ,2.0 s, and 5.0 s ? b. How far does it travel in the time interval t = 0 to t =3.0 s? a. What is its acceleration and velocity at times t =1.0 s ,2.0 s, and 5.0 s ? b. How far does it travel in the time interval t = 0 to t =3.0 s?Explanation / Answer
Accleration a = b t ^0.5 where b= 4.5 m / s ^ 2.5 velocity v = intergarl a dt = b [ t^1.5 / 1.5 ] + C = ( b / 1.5 ) t ^ 1.5 + C --( 1) where C = integration constant At time t = 0 , velocity v = 0 substitue this condition in eq ( 1) we get C = 0 So, eq ( 1) becomes , v = ( 4.5 / 1.5 ) t ^1.5 = 3 t ^ 1.5 (a). Accleration at t= 1 is a = 4.5 * 1 ^ 0.5 =4.5 m / s ^ 2 velocity at t= 1 is v = 3 t ^ 1.5 = 3 m / s Accleration at t = s is a ' = 4.5 ( 2 ) 0.5 = 4.5 * 1.414 = 6.36 m / s ^ 2 velocity at time t=- 2 s is v ' = 3 t ^ 1.5 = 3 ( 2 )^1.5 = 8.48 m / s accleration at t = 5 s is a " = 4.5 t ^0.5 = 4.5 * 5 ^ 0.5 = 10.06 m / s ^2 velocity at t= 5 s is v " = 3 t ^ 1.5 = 3 * 5 ^ 1.5 = 33.54 m / s (b).distance travelled = integral v dt with in thelimits t = 0 to t = 3 = 3 * t ^ 2.5 / 2.5 = ( 3 / 2.5 ) t ^ 2.5 with in th elimits t=0 tot=3 = 12.47 mRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.