a car moving on a straight road with a speed of 20m/s. at t=0 the driver of the car applies the breaks after watching an obstacle 150m ahead. after application of breaks, the car s with 2m/s^2. find the position of the car from the obstacle at t=15s.

Post it.

To find the position of the car at a specific time, we need to calculate the distance covered by the car during that time.

We can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance covered
u = initial velocity
t = time
a = acceleration

Given:
Initial velocity (u) = 20 m/s
Acceleration (a) = -2 m/s^2 (negative because the car is decelerating)
Time (t) = 15 s
Distance to obstacle (d) = 150 m

To find the position of the car from the obstacle at t = 15s, we'll first calculate the distance covered by the car during this time:

Using the equation of motion:
s = ut + (1/2)at^2

Substituting the given values:
s = (20 m/s)(15 s) + (1/2)(-2 m/s^2)(15 s)^2

Calculating the value of s:

s = 300 m - (1/2)(2 m/s^2)(225 s^2)
s = 300 m - 2(112.5 m)
s = 300 m - 225 m
s = 75 m

Therefore, the position of the car from the obstacle at t = 15s is 75 meters.

d = 20(15) -(1/2)(2)(15^2)

= 300 - 225 = 75 m
150 - 75 = 75