A motor car moving at a speed of 72 , km/h ca | vedclass

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(B) The initial speed of both the car and the truck is $u = 72 \, km/h = 72 	imes \frac{5}{18} = 20 \, m/s$.For the truck,the time to stop is $t_t =

Vedclass · Accepted Answer

$1.25 \, m$

Vedclass · Answer

(B) The initial speed of both the car and the truck is $u = 72 \, km/h = 72 	imes \frac{5}{18} = 20 \, m/s$.For the truck,the time to stop is $t_t = 5.0 \, s$. Using $v = u + a_t t$,we get $0 = 20 + a_t(5)$,so $a_t = -4 \, m/s^2$.For the car,the time to stop is $t_c = 3.0 \, s$. Using $v = u + a_c t$,we get $0 = 20 + a_c(3)$,so $a_c = -20/3 \, m/s^2$.Let $t$ be the time when the velocities of the car and truck become equal to avoid collision. The car has a reaction time of $0.5 \, s$,so it travels at constant speed for $0.5 \, s$ and then decelerates for $(t - 0.5) \, s$.The velocity of the truck at time $t$ is $v_t = 20 - 4t$.The velocity of the car at time $t$ is $v_c = 20 - (20/3)(t - 0.5)$.Setting $v_c = v_t$: $20 - (20/3)(t - 0.5) = 20 - 4t \Rightarrow (20/3)(t - 0.5) = 4t \Rightarrow 5(t - 0.5) = 3t \Rightarrow 2t = 2.5 \Rightarrow t = 1.25 \, s$.The distance covered by the truck in $t = 1.25 \, s$ is $s_t = 20(1.25) + 0.5(-4)(1.25)^2 = 25 - 3.125 = 21.875 \, m$.The distance covered by the car in $t = 1.25 \, s$ is $s_c = (20 	imes 0.5) + [20(1.25 - 0.5) + 0.5(-20/3)(1.25 - 0.5)^2] = 10 + [15 - 2.0833] = 22.9167 \, m$ (approx). Calculating precisely: $s_c = 10 + 20(0.75) - (10/3)(0.75)^2 = 10 + 15 - 1.875 = 23.125 \, m$.The required distance is $s_c - s_t = 23.125 - 21.875 = 1.25 \, m$.

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