# A race car speeds up as it enters a wide horizontal curve, going from 72 km/h to 108km/h in the 10s that it takes to round the curve. The radius of the curve is 400m. While in the curve, the car's speed is increasing at a constant rate.What is the car's tangential acceleration?At 108 km/h, find the centripetal acceleration and the total acceleration.

## Expert Answers

First, let us calculate the tangential acceleration of the car. This acceleration is responsible for the increase in speed. Since the tangential acceleration is constant, all we have to do is divide the change in the speed by the change in time:

(1) `a_t=(v_f - v_i)/(Delta t)`

But first, let's change our speed to the right units. We want to go from kilometers per hour to meters per second. To do this, we divide by `3.6` . Now, the final speed `v_f` is `30m//s` and the initial speed `v_i` is `20m//s` .

Now, using the definition (1), the tangential acceleration is, in m/s²:

`a_t= (30-20)/(10)=1 m//s^2`

This answers the first question.

Now, to calculate the centripetal acceleration at a given speed, we use the following formula:

(2) `a_c=v^2/r`

Plugging in `v=108km//h=30m//s` and `r=400m` , we find that the centripetal acceleration when the car is going at 108km/h is equal to:

`a_c=(30)^2/400=2.25m//s^2`

To calculate the total acceleration `a` when the car is at 108km/h, we have to add the centripetal acceleration to the tangential acceleration. We use the facts that both accelerations are vectors and that the tangential acceleration vector is perpendicular to the centripetal acceleration vector (the latter points to the center of the curve). So, to add them, we use the Pythagorean theorem:

(3) `a^2=a_c^2+a_t^2=2.25^2+1^2=6.0625`

Taking the square root, we find that:

`a=2.462m//s^2`

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