Question

the displacemant x of a particle varies with time as' x=4t^(2)-15t+25 .Find its velocity and acceleration at t=1 s.​

Answer 1

Answer:

Let us assume SI units are to be used

position x = ( 4t2 -15t + 25 ) m

velocity dx/dt = ( 8t - 15 ) m/s

acceleration = 8 m/s2

Since acceleration is constant, this motion is uniformly accelerated motion.

velocity will become zero when t = 15/8 seconds

at t = 15/8 s, when velocity becomes zero, position of particle = 4×(15/8)2 -15(15/8) +25 ≈ 11 m

Answer 2

Answer:

=> Let us assume SI unita are to be used position x = ( 4t2 - 15t + 25) m.

=> Velocity dx/dt = ( 8t - 15) m/s.

=> Acceleration = 8 m/s2.

=> Since acceleration is constant, this motion is uniformly accelerated motion.

=> Velocity become zero when t = 15/8 s.

=> at t = 15/8 s, when velocity becomes zero, position of particle = 4 × (15/8) 2 - 15 (15/8) + 25 = 11m.

Explanation:

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