Question 34:
(a) Derive the formula : v = u + at, where the symbols have usual meanings.
(b) A bus was moving with a speed of 54 km/h. On applying brakes it stopped in 8 seconds. Calculate the acceleration.
Lakhmir Singh Physics Class 9
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a). Let u be the initial velocity
v = Final velocity
In the velocity time graph , the particles increase its speed from u in t = 0 to in t = t
So , we can write expression for acceleration as
a = v - u/t - 0
a = ( v - u ) / t
at = v - u
v = u + at
b) Initial speed ( u ) = 54 km/hr = 54 *1000 m/ 3600 s = 54*5/18 = 15 m/s
( 1 km = 1000 m and 1 hr = 60 mins = 60*60 secs = 3600 secs )
Time = 8 secs .
Final velocity ( V ) = 0
We know that ,
v = u + at
0 = 15 + a*8
0 = 15 + 8a
a = -15/8 m/s²
Therefore , acceleration is -15/8 m/s²
also ,
S = ut + 1/2at²
S = 15*8² + 1/2* ( -15/8)*8²
= 15*64 - 15*4
= 15( 64 - 4 )
= 15(60)
S = 900 m
Hence stopping distance traveled before stopping is 900 m .
v = Final velocity
In the velocity time graph , the particles increase its speed from u in t = 0 to in t = t
So , we can write expression for acceleration as
a = v - u/t - 0
a = ( v - u ) / t
at = v - u
v = u + at
b) Initial speed ( u ) = 54 km/hr = 54 *1000 m/ 3600 s = 54*5/18 = 15 m/s
( 1 km = 1000 m and 1 hr = 60 mins = 60*60 secs = 3600 secs )
Time = 8 secs .
Final velocity ( V ) = 0
We know that ,
v = u + at
0 = 15 + a*8
0 = 15 + 8a
a = -15/8 m/s²
Therefore , acceleration is -15/8 m/s²
also ,
S = ut + 1/2at²
S = 15*8² + 1/2* ( -15/8)*8²
= 15*64 - 15*4
= 15( 64 - 4 )
= 15(60)
S = 900 m
Hence stopping distance traveled before stopping is 900 m .
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