EXERCISE 1
A particle travels along a straight line with a velocity of v = (4t-3t2) m/s, where t is in seconds. Determine the position of the particle when t= 4s. Consider x=0 when t=0.
SOLUTION:
Step 1: We know that:
\(v=4t-3t^2\)
Step 2: In order to obtain the position we use the reliable old
\(v=\frac{dx}{dt}\)
Step 3: Clearing dx
\(dx=v\ dt\)
Step 4: We substitute the value of v
\(dx=\left(4t-3t^2\right)dt\)
Step 5: Now we proceed to integrate with respect to t
Knowing the formula ∫ \(u^ndu\) we have to
\(\int_0^xdx=\int_0^t\left(4t-3t^2\right)dt\)
\(x=\int_0^t4t\ dt\ -\int_0^t3t^2dt\)
\(=4\frac{t^2}{2}\ \left|_{_0\ }^{^t}-3\frac{t^3}{3}\right|_{_0}^{^t}\)
\(=2t^2-t3\)
x is equal to: \(x\left(t=4\right)\)\(=2\left(4\right)^2-\left(4\right)^3\)
Step 6: Following the formula the final result is
\(=32-64\)\(=-32m\)
EXERCISE 2