Two cars 6

Two cars leave the same place one after the other 15 seconds apart. Both are moving with uniform acceleration and zero initial velocity, 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 in which the cars overtake,
b) the magnitude of the velocities of both cars at the moment of overtaking.

Correct 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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