Question

an object starts from rest and acclerates 3meter per second square for 4 seconds then it starts retarding for 2meter per second square till it comes to rest . what is the total distance covered in the process​

Answer 1

Answer:

60meter

Explanation:

distance cover in part one

here a is 3m/s²;t is 4second; u is 0

So

s= ut+1/2at²

s=0*4+3*4*4/2

s=3*4*2

s=24m

andv=u+at

v=0+3*4

v=12m/s

for second part of motion

here a is 2m/s²; v is 0 ;u is 12m/s

v²-u²=2as

s=v²-u²/2a

s=0²-12²/2*2

s=144/4

s=36m

total distance is 24+36=60m

Answer 2

Answer:18.25m

Explanation:u=0

v=3m/s

t=4s

a=-2m/s^2

S1=ut-1/2at^2[for retardation]

S1=0×t-1/2×(-2)×(4)^2

S1=0-(-16)

S1=16m

v^2=u^2+2aS2[for retardation]

(3)^2=0-2(-2)S2

9=4S2

S2=9/4m

Total distance S=S1+S2=9/4+16=(9+64)/4

S=73/4

S=18.25m