Two cars 6

Two cars leave the same place one after the other, 15 seconds apart. Both move with uniform acceleration from rest: the first car with an acceleration of 0.5 m/s², and the second car with an acceleration of 2 m/s².

Determine:
a) the time and distance at which the second car overtakes the first,
b) the speeds of both cars at the moment of overtaking.

Final Answer:

t =  30 s
s =  225 m
v₁ =  15 m/s
v₂ =  30 m/s

Step-by-step explanation:

$$\begin{gathered} a_1=0.5{\text{m/s}}^2 \\ {a}_2=2{\text{m/s}}^2 \\ {t}_1=15\text{s} \\ {s}_1=s_2 \\ \frac{1}{2}\cdot a_1\cdot t^2=\frac{1}{2}\cdot a_2\cdot (t-t_1{)}^2 \\ {a}_1\cdot x^2=a_2\cdot (x-t_1{)}^2 \\ 0.5\cdot x^2=2\cdot (x-15{)}^2 \\ -1.5x^2+60x-450=0 \\ 1.5x^2-60x+450=0 \\ a=1.5;b=-60;c=450 \\ D=b^2-4ac=60^2-4\cdot 1.5\cdot 450=900 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{60\pm \sqrt{900}}{3} \\ {x}_{1,2}=\frac{60\pm 30}{3} \\ {x}_{1,2}=20\pm 10 \\ {x}_1=30 \\ {x}_2=10 \\ t>t_1 \\ t=x_1=30=30\text{s} \end{gathered}$$

Our quadratic equation calculator calculates it.

$s=\frac{1}{2}\cdot a_1\cdot t^2=\frac{1}{2}\cdot 0.5\cdot 30^2=225\text{m}$

$v_1=a_1\cdot t=0.5\cdot 30=15\text{m/s}$

$v_2=a_2\cdot (t-t_1)=2\cdot (30-15)=30\text{m/s}$

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