Question

if x=t³+3t²-2t+2 find distance velocity and acceleration in 1 second​

Answer 1

Answer:

for distance

x = t³+3t²-2t+2

now at t = 1 second

x = (1)³ + 3(1)² - 2(1) + 2

x = 1 + 3 - 2 + 2

x = 4 unit

dx/dt = velocity

velocity = v

and x = t³+3t²-2t+2

now dx/dt = d/dt (t³+3t²-2t+2)

v = 3t² + 6t - 2

at t=1 second

v = 3(1)² + 6(1) - 2

v = 3 + 6 - 2

v = 7 unit / second

now acceleration

acceleration = dv/dt

acceleration = a

dv/dt = d/dt (3t² + 6t - 2)

a = 6t + 6

at t = 1 second

a = 6(1) + 6

a = 12 unit²/second

hope this will help you

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god bless you