I think Talal's result is right. We have a formula: v' = v + at => v' - v = at = .89 m/s^2 * 5s = 4.45 m/s With: v: initial speed; v': speed after 5s; a: acceleration; t: time; the change in speed is v'- v.
Taltal is correct; as the acceleration is constant, the velocity is therefore the product of time and acceleration, hence we receive that a*t=Δv or that .89*5=4.45m/s.
Welcome to the blog, Hamzah! I'm so glad that I finally found a topic that interests you! What an epiphany this has been for me! Maybe we should discuss physics in class instead of whatever we're doing. Then you can be the teacher and I can sit and text in class ;)
v = at = .89m/s^2 * 5s = 4.45m/s
ReplyDeleteShukran, Talal. I was helping my son with his homework and I wanted to make sure I was doing it right.
DeleteI think Talal's result is right.
ReplyDeleteWe have a formula: v' = v + at => v' - v = at = .89 m/s^2 * 5s = 4.45 m/s
With: v: initial speed; v': speed after 5s; a: acceleration; t: time; the change in speed is v'- v.
Taltal is correct; as the acceleration is constant, the velocity is therefore the product of time and acceleration, hence we receive that a*t=Δv or that .89*5=4.45m/s.
ReplyDeleteHey, guys. The question is "the change in the speed". So the answer is a(=v/s). So the answer is 0.89 m/s^2.
ReplyDeleteIf it started to move from rest, then the final speed is 4.45 m/s
ReplyDeleteaccording to this fomula : Vf - Vi = a t
Welcome to the blog, Hamzah! I'm so glad that I finally found a topic that interests you! What an epiphany this has been for me!
DeleteMaybe we should discuss physics in class instead of whatever we're doing. Then you can be the teacher and I can sit and text in class ;)
Talal , Hang and Hamzah are all right . Great.
ReplyDelete