Question

How to solve the following (Images attached)

Answer 1

What level of math are you in?

For the first question, if you're familiar calculus, the equation for the velocity can be found by taking the first derivative of the equation for the position. The acceleration is the second derivative of the position equation.

That is to say, if s(t) is your distance, then:
\[\mathrm{V(t)} = \frac{d\,s(t)}{dt}\]
\[\mathrm{a(t)} = \frac{d],v(t)}{dt} = \frac{d^2\,s(t)}{dt^2}\]

If not, then you should take a look at the kinematic equations for constant acceleration. Since your position equation is quadratic in t, the acceleration is constant, and will fit these.

Answer 2

Bleh, equation problems.

\[\mathrm{v(t)} = \frac{d\,\mathrm{s(t)}}{dt}\]
\[\mathrm{a(t)} = \frac{d\,\mathrm{v(t)}}{dt} = \frac{d^2\,\mathrm{s(t)}}{dt^2}\]

Much better :)

Answer 3

You can do something similar with the second question. Either use the kinematic equations (calculate the acceleration, and use the initial position and velocity), or calculate the acceleration and integrate once for velocity, and twice for the position (the integrating constants are your initial conditions. That is, the integrating constant when you integrate the first time is the initial velocity. The constant after the second integral is the initial position).

Answer 4

Thank You

Answer 5

Can you work out the answers for the second question. Just would like to know if i am going right..

Answer 6

\[v(t) = 0.1t\]
\[v(0) = 0\]
\[R(t) = \int\limits v(t) dt\]\[R(0) = -7\]

So, for part i:
\[R(10) = \int0.1t\, dt = (0.05\,t^2 -7 )|_{t=10} = -2\]
And for part ii:
R(5)-R(3) = \int_3^5 = 0.05(5^2-3^2) = 0.8

Answer 7

Hmmm, second equation didn't parse. The math environment disappeared, somehow.

\[\int\limits_3^5 0.1t dt = 0.05(5^2-3^2) = 0.8\]

Answer 8

Thank you very much. Really Appreciate your help.