The car starts at rest and accelerates uniformly with magnitude [tex]a[/tex], so the distance [tex]x[/tex] it travels in time [tex]t[/tex] is
[tex]x = \dfrac12 at^2[/tex]
Solve for [tex]a[/tex].
[tex]300\,\mathrm m = \dfrac12 a (10\,\mathrm s)^2 \implies a = \dfrac{600\,\rm m}{100\,\mathrm s^2} = \boxed{6\dfrac{\rm m}{\mathrm s^2}}[/tex]
By Newton's second law, the magnitude of force [tex]F[/tex] acting on the object, if its mass [tex]m[/tex] is 1500 kg, is
[tex]F = ma = (1500\,\mathrm{kg})\left(6\dfrac{\rm m}{\mathrm s^2}\right) = \boxed{9000\,\mathrm N} = 9\,\mathrm{kN}[/tex]
