a body starting from rest and travelling with uniform acceleration has a velocity of 40m/s after 10second at A. velocity of the body 4second before it crosses the point A is
Answers
Given,
initial velocity,u= 0m/s
(since the body is starting from rest)
Final velocity (v)= 40m/s
Time = 10seconds
accelearation(a)=?
Now, by using the kinematics equations=>
v = u + at
=>40 = 0 + a×10
=>a = 4m/s^2
Hence according to the question we have=>
4 seconds before it crosses A i.e
10-4 = 6 seconds after the body is in motion.
Hence,by using Equations of kinematics=>
v= u + at
=>v = 0 + 4×(6)
=>V = 24 m/s.
Hence the speed of the body 4 seconds before crossing point A is 24m/s.
Remember
Three equations of kinematics =>
1)V = U + a
2)V^2 - U^2 = 2as
3) s= ut+ (1/2) × (a × t^2)
Answer:
Explanation:
Answers
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generalRd
GeneralRdGenius
Given,
initial velocity,u= 0m/s
(since the body is starting from rest)
Final velocity (v)= 40m/s
Time = 10seconds
accelearation(a)=?
Now, by using the kinematics equations=>
v = u + at
=>40 = 0 + a×10
=>a = 4m/s^2
Hence according to the question we have=>
4 seconds before it crosses A i.e
10-4 = 6 seconds after the body is in motion.
Hence,by using Equations of kinematics=>
v= u + at
=>v = 0 + 4×(6)
=>V = 24 m/s.
Hence the speed of the body 4 seconds before crossing point A is 24m/s.