A car deccelerates from a speed of 20m/s to a rest in a distance of 100m.What was its acceleration,assumed constant?
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3
initial velocity (u) = 20m/s
final velocity (v) = 0m/s
distance (s) = 100m
acceleration (a) = ?
using third equation of motion ,
v² = u²+2as
0² = (20)²+2(a)(100)
0 = 400 + 200a
-400 = 200a
a = -400/200 => -2 m/s²
negative sign denotes retardation of the body
hope this helps
final velocity (v) = 0m/s
distance (s) = 100m
acceleration (a) = ?
using third equation of motion ,
v² = u²+2as
0² = (20)²+2(a)(100)
0 = 400 + 200a
-400 = 200a
a = -400/200 => -2 m/s²
negative sign denotes retardation of the body
hope this helps
Answered by
0
V²-U² = 2aS
V=0
-400 = 2×a× 100
a = -2m/sec²
hope this helps.
V=0
-400 = 2×a× 100
a = -2m/sec²
hope this helps.
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