A particle starts to move in a straight line from a point with velocity 10m/s and acceleration -2m/s^2. Find the position and velocity of the particle at:a. 5 sec and 10 sec.
Answers
Answer:
Displacement at t = 10 s is
Displacement at t = 10 s iss
Displacement at t = 10 s iss ′
Displacement at t = 10 s iss ′ =ut
Displacement at t = 10 s iss ′ =ut ′
Displacement at t = 10 s iss ′ =ut ′ +
Displacement at t = 10 s iss ′ =ut ′ + 2
Displacement at t = 10 s iss ′ =ut ′ + 21
Displacement at t = 10 s iss ′ =ut ′ + 21
Displacement at t = 10 s iss ′ =ut ′ + 21 at
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 2
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10)
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s is
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+at
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1 .
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1 .i.e., velocity is 10ms
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1 .i.e., velocity is 10ms −1
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1 .i.e., velocity is 10ms −1 opposite to the direction of initial starting velocity.
Displacement at t = 10 s iss ′ =ut ′ + 21 at ′2 =10×10+ 21 ×(−2.0)×(10) 2 =100−100=0(zero)i.e., after 10 s, the particle will come back to the starting point.Velocity at t = 10 s isv=u+atv=10+(−2)∗10=−10ms −1 .i.e., velocity is 10ms −1 opposite to the direction of initial starting velocity.Answere by Swayamprabha Nayak