a car moving with constant acceleration covers the distance between two points 58.0m apart in 6.20s it's speed as it passes the second point is 15.0
Answers
Equation of motion,v = u + at
Equation of motion,v = u + ats = ut + 1/2 a t^2
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2as
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/s
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocity
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = acceleration
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60m
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6s
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 6
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 60
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 609u^2 - 720u + 3600 - 2025 = 0
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 609u^2 - 720u + 3600 - 2025 = 0u^2 - 40u + 175 = 0
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 609u^2 - 720u + 3600 - 2025 = 0u^2 - 40u + 175 = 0Hence,(u -5)(u - 35) = 0
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 609u^2 - 720u + 3600 - 2025 = 0u^2 - 40u + 175 = 0Hence,(u -5)(u - 35) = 0Only possible value for u = 5 m/s
Equation of motion,v = u + ats = ut + 1/2 a t^2v^2 = u^2 + 2asWhere v = final velocity = 15 m/su = initial velocitya = accelerations = distance covered = 60mt = time of motion = 6sHence we have, 60 = u * 6 + 1/2 a * 6 * 660 - 6u = 18a15 * 15 = u^2 + 2a * 60225 = u^2 + 2 * (60 - 6u)/18 * 609u^2 - 720u + 3600 - 2025 = 0u^2 - 40u + 175 = 0Hence,(u -5)(u - 35) = 0Only possible value for u = 5 m/s