+502 A car travelling at 90 km/h is 500 m behind another car travelling at 70 km/h in the same direction. How long will it take the first car to catch the second?
+26393 A car travelling at 90 km/h is 500 m behind another car travelling at 70 km/h in the same direction.
How long will it take the first car to catch the second?
\(\begin{array}{cll} \text{Let } v_1 &=& 90\ \frac{km}{h} \\ \text{Let } v_2 &=& 70\ \frac{km}{h} \\ \text{Let } d = 500\ m &=& 0.5\ km \\ \text{Let } t &=& time \\ \end{array}\)
if d is the distance:
\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& v_1\cdot t - 0.5 \\ d_{\text{car}_2} &=& v_2\cdot t \\ \hline \end{array}\)
The first car catch the second if \(d_{\text{car}_1} = d_{\text{car}_2}\)
\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& d_{\text{car}_2} \\ v_1\cdot t -0.5\ km &=& v_2\cdot t \quad & | \quad -v_2\cdot t +0.5 \\ v_1\cdot t - v_2\cdot t &=& 0.5\ km \\ t\cdot ( v_1 - v_2 ) &=& 0.5\ km \quad & | \quad :( v_1 - v_2 ) \\ t &=& \frac {0.5\ km} { v_1 - v_2 } \quad & | \quad v_1 = 90\ \frac{km}{h} \qquad v_2 = 70\ \frac{km}{h} \\ t &=& \frac {0.5\ km} { 90\ \frac{km}{h} - 70\ \frac{km}{h} } \\ t &=& \frac {0.5\ km} { 20\ \frac{km}{h} } \\ t &=& \frac {0.5} { 20 }\ h \\ t &=& 0.025\ h \quad & | \quad 1\ h = 60\min.\\ t &=& 0.025\cdot 60\min.\\ \mathbf{t} & \mathbf{=} & \mathbf{1.5\min.} \\ \hline \end{array}\)
+26393 A car travelling at 90 km/h is 500 m behind another car travelling at 70 km/h in the same direction.
How long will it take the first car to catch the second?
\(\begin{array}{cll} \text{Let } v_1 &=& 90\ \frac{km}{h} \\ \text{Let } v_2 &=& 70\ \frac{km}{h} \\ \text{Let } d = 500\ m &=& 0.5\ km \\ \text{Let } t &=& time \\ \end{array}\)
if d is the distance:
\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& v_1\cdot t - 0.5 \\ d_{\text{car}_2} &=& v_2\cdot t \\ \hline \end{array}\)
The first car catch the second if \(d_{\text{car}_1} = d_{\text{car}_2}\)
\(\begin{array}{|rcll|} \hline d_{\text{car}_1} &=& d_{\text{car}_2} \\ v_1\cdot t -0.5\ km &=& v_2\cdot t \quad & | \quad -v_2\cdot t +0.5 \\ v_1\cdot t - v_2\cdot t &=& 0.5\ km \\ t\cdot ( v_1 - v_2 ) &=& 0.5\ km \quad & | \quad :( v_1 - v_2 ) \\ t &=& \frac {0.5\ km} { v_1 - v_2 } \quad & | \quad v_1 = 90\ \frac{km}{h} \qquad v_2 = 70\ \frac{km}{h} \\ t &=& \frac {0.5\ km} { 90\ \frac{km}{h} - 70\ \frac{km}{h} } \\ t &=& \frac {0.5\ km} { 20\ \frac{km}{h} } \\ t &=& \frac {0.5} { 20 }\ h \\ t &=& 0.025\ h \quad & | \quad 1\ h = 60\min.\\ t &=& 0.025\cdot 60\min.\\ \mathbf{t} & \mathbf{=} & \mathbf{1.5\min.} \\ \hline \end{array}\)
+37155 Rate times time = distance the distance is 500m (.5 km)
and the rate is the net speed difference (90-70) = 20 kmh
20 x t = .5 km
t = .5/20 = .025 hr = 1.5 minutes
