A car can decelerate at -3.36-m/s2 without skidding when coming to rest on a level road. What would its deceleration (in m/s2) be if the road is inclined at 8.6o and the car moves uphill? (Assume the same static friction coefficient.

on the incline

friction=mu*mg*costheta
but mu*mg=m*3.36 or mu=3.36/g
weight down hill: mg*sinTheta
net force down hill= weightdown+friction
= mg*sinTheta+3.36/g*mg*costheta
=mg(sinTheta+3.36costheata/g)

so acceleration would be net force/mass down the hill would be
g*sintheta+3.36cosTheta/g and if you want the deacceleration up the hill, it is the negative of that.

I keep getting the wrong answer. i must be putting it in my calculator wrong.

To find the deceleration of the car when moving uphill on an inclined road, we need to consider the forces acting on the car.

When the car is on a level road, the only force acting on it is the force of friction between the tires and the road surface. This frictional force provides the deceleration of -3.36 m/s^2.

When the road is inclined at an angle of 8.6 degrees, there are two forces acting on the car: the force of friction and the gravitational force pulling the car downhill. The component of the gravitational force parallel to the road will act to increase the deceleration, while the component of the gravitational force perpendicular to the road will act normal to the road surface and will not affect the deceleration.

To calculate the deceleration, we need to find the downward force component acting parallel to the road. This force component can be found using trigonometry.

The component of the gravitational force parallel to the road can be calculated using the formula:

Force_parallel = m * g * sin(theta)

Where:
m is the mass of the car
g is the acceleration due to gravity (approximately 9.8 m/s^2)
theta is the angle of inclination in radians

Now let's substitute the given values into the formula:

theta = 8.6 degrees
m and g are not given, but they are not required for our calculation

First, we need to convert the angle from degrees to radians:

theta_rad = theta * (pi / 180)
theta_rad = 8.6 * (pi / 180)
theta_rad = 0.1495 radians

Now we can calculate the force parallel to the road:

Force_parallel = m * g * sin(theta_rad)

Since we are assuming that the static friction coefficient remains the same, we can equate the force of friction to the parallel component of the gravitational force:

Force_friction = m * a

Where:
a is the deceleration

Setting the force of friction equal to the parallel component of the gravitational force:

m * a = m * g * sin(theta_rad)

We can cancel out the mass:

a = g * sin(theta_rad)

Now, substitute the given values:

a = 9.8 * sin(0.1495)
a ≈ 1.37 m/s^2

Therefore, the deceleration of the car when moving uphill on an inclined road would be approximately 1.37 m/s^2.