Answer:
88.33 m
Explanation:
For AB:
u = 0, t = 5 s, a = 2 m/s^2
Let the distance between A and B be s1 and the velocity at B is v.
Use second equation of motion
[tex]s_{1}=ut + 1/2 at^{2}[/tex]
[tex]s_{1}=0 + 1/2 \times 2\times 5 \times 5[/tex]
s1 = 25 m
Use first equation of motion
v = u + a t
v = 0 + 2 x 5 = 10 m/s
For BC:
Let the distance from B to C is s2.
time = 3 s,
speed, v = 10 m/s
s2 = v x t = 10 x 3 = 30 m
For CD:
final velocity, v' = 0
initial velocity, v = 10 m/s
acceleration, a = - 1.5 m/s^2
Let the distance from C to D is s3.
Use third equation of motion
[tex]v'^{2}=v^{2}+2as_{3}[/tex]
[tex]0^{2}=10^{2}-2 \times1.5 \times s_{3}[/tex]
s3 = 33.33 m
The total distance covered is s
s = s1 + s2 + s3 = 25 + 30 + 33.33 = 88.33 m
