A car is moving on a straight road with a speed of 20 m/s .At t = 0 ,the driver of car applies the brakes after watching and obstacle 150 m ahead .After application of brakes ,the car retards with 2m/s2 .Find the position of car from the obstacle at t = 15 sec

Here initial velocity u=20m/s,Final velocity v=0,Acceleration a=

Now,To calculate the time taken by car to stop

v=u+at

0=20+(-2)t

t=10sec

Car will stop in 10 seconds,it means after 15 sec car would have the same position as at the end of 10 seconds.

The distance travelled by the car in 10 seconds

The position of car from the obstacle=150-100=50m.

  • 34

Ok

For Finding The Position We Use Second Equation Of Motion

Here u = 20,a = -2,t = 15

s = ut+1/2*a*t*t

s = 20*15-1/2*2*15*15

s = 300-225 = 75m

So The Car Would Be 75 m From The Obstacle

  • -23

No it's answer is 50 m ahead of obstacle 

by v = u - at

 0 = 20 - 2 t

 t = 10 sec

car stops in 10 sec

so position of car 

s = ut - 1/2 at2

 s = 20 * 10 - 1/2 * 2 * 100

s = 100 m from initial point

so 50 m ahead of obstacle

  • 7
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