Answer:
(a) [tex]0.984781 m/s^{2}[/tex]
(b) 157.76679 m
(c) 17.62757 m/s
Explanation:
Net force acting, [tex]F_{net}=2.96*10^{3}-1.86*10^{3}=1.1*10^{3}N[/tex]
Acceleration, [tex]a=\frac {F_{net}}{m}[/tex] where m is the mass
[tex]a=\frac {1.1*10^{3}}{1117}= 0.984781 m/s^{2}[/tex]
(b)
From kinematic equation
[tex]s=ut+0.5at^{2}[/tex] where s is displacement, u is initial velocity, t is time taken and a is acceleration.
Considering that u=0 since it starts at rest
[tex]s=0.5at^{2}[/tex] and substituting the value of a as calculated in part a, t is 17.9 s hence
[tex]s=0.5*0.984781*17.9^{2}=157.76679 m[/tex]
(c
)
From kinematic equation
v=u+at where u and v are initial and final velocities respectively, a is acceleration and t is duration in seconds. Since it starts from rest, u=0 and substituting the value of a as found in part a, t given as 17.9 s we get
V=0+0.984781*17.9=17.62757 m/s
