<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em></em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em></em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em></em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em></em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em></em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em></em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em></em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em></em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
From the basic "heat lost by hot object=heat gained by colder object" principle, we have m1c1ΔT1=m2c2ΔT2 where m1= 1kg m2=4kg c1=900J/kg k c2=4200J/kg k With this information at hand we have m1c1(90-T)=m2c2(T-25) after substituting the given values we can find that T=28.3^{0}c
