Question

✨QuEStiON for mods and Brainly star :-✨

Long Answer type Question


A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a
signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5 s .​

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

➳ Time taken by car to stop = 3s

➳ Time taken by truck to stop = 5s

➳ Both are moving at a speed of 72km/h (20m/s).

➳ Human (driver of car) response time = 0.5s

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

፨ Minimum distance required for the car to be distant from the truck to avoid accident.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

⛥ Retardation of car :

After applying brakes, car stops after 3s.

$:\implies\tt\:v=u+a_ct$

$:\implies\tt\:0=20+a_c(3)$

$:\implies\bf\:a_c=-6.67\:ms^{-2}$

⛥ Retardation of truck :

After applying brakes, truck stops after 5s.

$:\implies\tt\:v=u+a_tt$

$:\implies\tt\:0=20+a_t(5)$

$:\implies\bf\:a_t=-4\:ms^{-2}$

➢ Let initial distance b/w car and truck be s. (at that instant when truck gives the signal)

➢ To avoid the collision, velocity of the car and truck after time t should be equal.

⛥ Velocity of car after time t :

$:\implies\tt\:v_c=u+a_c(t-0.5)$

Because, driver applies brakes after 0.5s.

$:\implies\bf\:v_c=20-6.67(t-0.5)$

⛥ Velocity of truck after time t :

$:\implies\tt\:v_t=u+a_tt$

$:\implies\bf\:v_t=20-4t$

❍ After time t :

$:\implies\tt\:v_c=v_t$

$:\implies\tt\:20-6.67(t-0.5)=4t$

$:\implies\tt\:12t=20t-10$

$:\implies\tt\:8t=10$

$:\implies\bf\:t=1.25\:s$

⛥ Distance covered by truck in 1.25s :

$:\implies\tt\:d_t=ut+\dfrac{1}{2}a_tt^2$

$:\implies\tt\:d_t=(20\times 1.25)+\dfrac{1}{2}(-4)(1.25)^2$

$:\implies\tt\:d_t=25-3.125$

$:\implies\bf\:d_t=21.875m$

⛥ Distance covered by car in 1.25s :

$:\implies\tt\:d_c=d_{in\:0.5s}+d_{in\:0.75\:s}$

$:\implies\tt\:d_c=(ut')+(ut+\dfrac{1}{2}a_c{t}^2)$

$:\implies\tt\:d_c=(20\times 0.5)+(20\times 0.75+\dfrac{1}{2}(-6.67)(0.75)^2)$

$:\implies\tt\:d_c=10+15-1.875$

$:\implies\bf\:d_c=23.125m$

◈ Therefore, minimum distance required b/w the car and truck to avoid accident = 23.125 - 21.875 = 1.5m :D

Nice Question (✊)