a car starts from rest and moves along x axis with constant acceleration of 5m/s2 for 8 seconds. if it then continue with constant velocity what distance will the car cover in 12 seconds since it stared from rest
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given for first 8 sec
u = 0
a = 5
t = 8
by using v= u + at
v = 0 + 5×8
v at 8th sec = 40m/ s
therefore distance covered in first 8 sec
= ut + 1/2 a t^2
= 0 + 1/2×5×8×8
= 5× 4×8
= 160m
then the distance covered in 8 to 12 sec
t = 4 sec
v = 40m/s
therefore distance = v× t
= 40×4
= 160 m
therefore total distance covered in 12 sec = 160+160
=320m
u = 0
a = 5
t = 8
by using v= u + at
v = 0 + 5×8
v at 8th sec = 40m/ s
therefore distance covered in first 8 sec
= ut + 1/2 a t^2
= 0 + 1/2×5×8×8
= 5× 4×8
= 160m
then the distance covered in 8 to 12 sec
t = 4 sec
v = 40m/s
therefore distance = v× t
= 40×4
= 160 m
therefore total distance covered in 12 sec = 160+160
=320m
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