Asked by Akshita
A body is moving with uniform acceleration describe 40m in first 5second and 65m in next 5second. It's initial velocity will be?
Answers
Answered by
Steve
You have
5v + a/2 * 5^2 = 40
10v + a/2 * 10^2 = 105
Now just solve for v.
5v + a/2 * 5^2 = 40
10v + a/2 * 10^2 = 105
Now just solve for v.
Answered by
MD SHAHNAWAZ
Help full for problems
Answered by
Abhijeet Sharma
5v + a/2 * 5^2 = 40 ____ (i)
10v + a/2 * 10^2 = 10^2 ____ (ii)
for (i) :
80 = 10v +25a
16 = 2v +5a _____ (a)
for (ii) :
105 = 10v + 100a/2
105 = 10v + 50a
21 = 2a + 10a _____ (b)
solve for( a )&( b )
we get ,
5a = 5
a=1
therefore :
2v +5=16
2v = 11
v = 5.5m/s
10v + a/2 * 10^2 = 10^2 ____ (ii)
for (i) :
80 = 10v +25a
16 = 2v +5a _____ (a)
for (ii) :
105 = 10v + 100a/2
105 = 10v + 50a
21 = 2a + 10a _____ (b)
solve for( a )&( b )
we get ,
5a = 5
a=1
therefore :
2v +5=16
2v = 11
v = 5.5m/s
Answered by
rakul kapoor
If initial velocity be u and the acceleration be a then the distance in first 5s will be S
5
=u(5)+
2
a(5)
2
=40m........(1)
and the distance in first 10s will be S
10
=u(10)+
2
a(10)
2
so the distance between time t=5s to t=10 will be
S
5
′
=S
10
−S
5
=65m ...........(2)
putting corresponding values and solving the equation-1 and equation-2 we get u=5.5m/s
Option C is correct.
5
=u(5)+
2
a(5)
2
=40m........(1)
and the distance in first 10s will be S
10
=u(10)+
2
a(10)
2
so the distance between time t=5s to t=10 will be
S
5
′
=S
10
−S
5
=65m ...........(2)
putting corresponding values and solving the equation-1 and equation-2 we get u=5.5m/s
Option C is correct.
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