Discussion Forum

Consider 'a' vehicle moving with a velocity 54 km/h. at a distance of 400m, from the traffic light brakes are applied. The acceleration of the vehicle, after the application of brakes, is $-0.3 m/s^{2}$. The vehicle's  position relative to the traffic light is 

Answer:

Explanation:

Given, initial velocity,

$u=54 km/h=54 \times \frac{5}{18}$m/s

 u=15 m/s

 Distance of signal from vehicle , d=400 m

 acceleration , a= -0.3m/s²

 Now, distance covered  by vehicle  , when it stops  is given by $v^{2}=u^{2}+2as$

  $0=(0.5)^{2}-2 \times 0.3 \times s \Rightarrow s= \frac{225}{0.6}=375m$

 Relative position =400-375=25m