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1] A Canadian sports car dealer claims that his product willaccelerate at a cons

ID: 1763779 • Letter: 1

Question

1] A Canadian sports car dealer claims that his product willaccelerate at a constant rate from rest to a speed of 110km/h in9.00s. What is the speed after the first 5.00s of accerelation inm/s? b] Awater balloon is released from rest from the top of theThurgood Marshall School of law. Neglecting the air resistance.Compute the velocity and position of the water balloon after 1.00s,2.00s and 3.00s. 1] A Canadian sports car dealer claims that his product willaccelerate at a constant rate from rest to a speed of 110km/h in9.00s. What is the speed after the first 5.00s of accerelation inm/s? b] Awater balloon is released from rest from the top of theThurgood Marshall School of law. Neglecting the air resistance.Compute the velocity and position of the water balloon after 1.00s,2.00s and 3.00s.

Explanation / Answer

           Given that the initial velocity is u = 0            Final velocity is v = 110 km/h = 30.55 m/s            Time taken is t1 = 9.00s ----------------------------------------------------------------------- (a) The acceleration is a = ( v -u ) /t1                                        = ( 30.55m/s -0 ) / 9.00s                                        = 3.39m/s2    Then the velocity at time t = 5.0s is V = u +at                                                          = 0 + (3.39 m/s2 )*5.0s                                                          = 16.97 m/s   (b) The velocity after time t is V1 = u + gt                                                   = 0+(9.8m/s2 )(1.0s)                                                   = 9.8 m/s The velocity after time t = 2.0is V1 = u + gt                                                      = 0 +(9.8m/s2 )(2.0s)                                                           = 19.6 m/s          The velocity aftertime t = 3.0 is V1 = u + gt                                                      = 0 +(9.8m/s2 )(3.0s)                                                           = 29.4 m/s

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