Question

please answer the questions​

Answer 1

Answer :

22 m/s²

Note :

• If x(t) is the expression for displacement (position) which is a function of time then the velocity is given by : v(t) = Dx(t) where D is the differential operator d( )/dt .

• If v(t) is the expression for the velocity which is a function of time then the acceleration is given by : a(t) = Dv(t) where D is the differential operator d( )/dt .

Solution :

• Given : Displacement, x(t) = (3t³+2t²+t-4)m

• To find : Acceleration at t = 1 sec

We have the displacement of the particle as -

x(t) = (3t³+2t²+t-4)m

Now ,

The velocity of the particle will be :

=> v(t) = Dx(t)

=> v(t) = d[x(t)]/dt

=> v(t) = d(3t³+2t²+t-4)/dt

=> v(t) = 3dt³/dt + 2dt²/dt + dt/dt - d4/dt

=> v(t) = 3•3t² + 2•2t + 1 - 0

=> v(t) = 9t² + 4t + 1

Now ,

The acceleration of the particle will be :

=> a(t) = Dv(t)

=> a(t) = d[v(t)]/dt

=> a(t) = d(9t² + 4t + 1)/dt

=> a(t) = 9dt²/dt + 4dt/dt + d1/dt

=> a(t) = 9•2t + 4•1 + 0

=> a(t) = 18t + 4

Now ,

The acceleration of the particle at t = 1 sec will be given as ;

a(1) = 18•1 + 4 = 22

Hence ,

The required answer is 22 m/s² .

Answer 2

Answer : Option B

• Here we are given the equation of the displacement of the body.
• The derivative of displacement gives us velocity. Then we differentiate again to obtain the equation of acceleration
• Basically we are obtaining the double derivative of position to find acceleration.

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Formulas used :

1. derivative of a constant is 0
2. d(a^n)/dx = na^(n-1)