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.