A car is moving with a velocity of 12 m/s. on applying brakes, it comes to rest in 6 seconds. Calculate the (1) acceleration (2) distance covered in this time period.
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0=12^2-2as , s= 72-18a ,144-2a(72-18a) it is a quadratic equation a is -ve find a then find s please mark brainliest
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Initial Velocity( u) = 12 m/s
Final Velocity ( v) = 0 m/s
It comes to rest in 6 sec
To find : a= ?
distance covered in this time period = ?
Solution
v = u + at
0 = 12 + a× 6
-12/6= a
a= -2 m/sec²
distance
( I ) s= ut +1/2 at²
(II) 2as = v²-u²
by II formula
2×(-2) × s= (0)²-(12)²
-4s= -144
s= 36 m
by l formula
s = (0)×6+1/2×(2)(6)²
s= 0+1(36)
s=36 m
Final Velocity ( v) = 0 m/s
It comes to rest in 6 sec
To find : a= ?
distance covered in this time period = ?
Solution
v = u + at
0 = 12 + a× 6
-12/6= a
a= -2 m/sec²
distance
( I ) s= ut +1/2 at²
(II) 2as = v²-u²
by II formula
2×(-2) × s= (0)²-(12)²
-4s= -144
s= 36 m
by l formula
s = (0)×6+1/2×(2)(6)²
s= 0+1(36)
s=36 m
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