At an accident scene, 276-m long skid marks are found on the road. If the car that made these skid marks can decelerate at –9.8 m/s2, how fast was the car going before braking?
To find the initial speed of the car before braking, we can use the following equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s since the car comes to a stop)
u = initial velocity (the speed we want to find)
a = acceleration (deceleration in this case, which is -9.8 m/s^2)
s = distance traveled (the length of the skid marks, which is 276 m)
Plugging in the values into the equation, we have:
0^2 = u^2 + 2(-9.8)(276)
Simplifying further:
0 = u^2 - 2(9.8)(276)
0 = u^2 - 5378.4
Rearranging the equation:
u^2 = 5378.4
Taking the square root of both sides to solve for u:
u = √5378.4
u ≈ 73.37 m/s
Therefore, the car was going approximately 73.37 m/s before braking.
To find out how fast the car was going before braking, we can use the equation of motion for constant deceleration.
The equation is:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 in this case because the car stops)
u = initial velocity (what we need to find)
a = acceleration (given as -9.8 m/s^2)
s = distance (given as 276 m)
Plugging in the values into the equation, we get:
0 = u^2 + 2(-9.8)(276)
Simplifying the equation:
0 = u^2 - 5392.8
Now, we can solve for u by taking the square root of both sides of the equation:
u^2 = 5392.8
u ≈ ±73.5119
Since we are interested in the speed of the car, we take the positive value:
u ≈ 73.5119 m/s
Therefore, the car was going approximately 73.5119 m/s before braking.