Answer:
(a) d = 20 m
(b) d' = 80 m
(c) x = 28 m
(d) x' = 96 m
Solution:
As per the question:
Initial velocity of the object, v = 0
Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]
(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:
[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]
(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:
[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]
Now, when initial velocity, v = 4 m/s, then
(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]
(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]
