Answer
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Hint: In this problem the two cars moves in the same direction with the velocity $60km{h^{ - 1}}$ and $70km{h^{ - 1}}$ respectively as shown in the figure below and the deceleration of the car $B$ is given as $20km{h^{ - 2}}$ we need to use the formula $s = ut + \dfrac{1}{2}a{t^2}$ to find the acceleration and time.
Complete step-by-step solution:
Given
Speed of car $A = 60km{h^{ - 1}}$
Speed of car $B = 70km{h^{ - 1}}$
Distance between car $B$ and $A = 2.5km$
deceleration, $a = - 20km{h^{ - 2}}$
For car $A$
${s_1} = ut = 60t$ ………..$\left( 1 \right)$
For car $B$
${s_2} = ut + \dfrac{1}{2}a{t^2}$
Substituting the given value we get
${s_2} = 70t + \dfrac{1}{2}\left( { - 20} \right){t^2}$
${s_2} = 70t - 10{t^2}$ …………$\left( 2 \right)$
From the diagram it is clear that $2.5 = {s_2} - {s_1}$ ……….$\left( 3 \right)$
Substituting equation $\left( 1 \right)$ and equation $\left( 2 \right)$ in equation $\left( 3 \right)$
$2.5 = 70t - 10{t^2} - 60t$
On simplifying
$10{t^2} - 10t + 2.5 = 0$
On factorization the above equation we get
$t = 0.5h$
Substituting in equation $\left( 2 \right)$
${s_2} = 70\left( {0.5} \right) - 10{\left( {0.5} \right)^2}$
${s_2} = 32.5km$
Hence the distance is $32.5km$ and time is $t = 0.5h$
Note: Since the speed of two cars are given in terms of $km{h^{ - 1}}$ and distance in $km$ we can do calculations by considering the same unit and after calculation we will get time $h$ and distance in $km$ .
Complete step-by-step solution:
Given
Speed of car $A = 60km{h^{ - 1}}$
Speed of car $B = 70km{h^{ - 1}}$
Distance between car $B$ and $A = 2.5km$
deceleration, $a = - 20km{h^{ - 2}}$
For car $A$
${s_1} = ut = 60t$ ………..$\left( 1 \right)$
For car $B$
${s_2} = ut + \dfrac{1}{2}a{t^2}$
Substituting the given value we get
${s_2} = 70t + \dfrac{1}{2}\left( { - 20} \right){t^2}$
${s_2} = 70t - 10{t^2}$ …………$\left( 2 \right)$
From the diagram it is clear that $2.5 = {s_2} - {s_1}$ ……….$\left( 3 \right)$
Substituting equation $\left( 1 \right)$ and equation $\left( 2 \right)$ in equation $\left( 3 \right)$
$2.5 = 70t - 10{t^2} - 60t$
On simplifying
$10{t^2} - 10t + 2.5 = 0$
On factorization the above equation we get
$t = 0.5h$
Substituting in equation $\left( 2 \right)$
${s_2} = 70\left( {0.5} \right) - 10{\left( {0.5} \right)^2}$
${s_2} = 32.5km$
Hence the distance is $32.5km$ and time is $t = 0.5h$
Note: Since the speed of two cars are given in terms of $km{h^{ - 1}}$ and distance in $km$ we can do calculations by considering the same unit and after calculation we will get time $h$ and distance in $km$ .
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