A car is moving along a circular path of radius 300 m with a speed of 30 m/s. If atsome instant, its speed increases at the rate of 4m/s2, then at that instant, find themagnitude of resultant acceleration

The resultant acceleration of a body moving in a circular path is given by the formula:

a = sqrt((r * w^2)^2 + (dw/dt)^2)

where:

• r is the radius of the circular path,
• w is the angular velocity, and
• dw/dt is the rate of change of angular velocity (also known as angular acceleration).

First, we need to convert the linear velocity to angular velocity. The relationship between linear velocity (v) and angular velocity (w) is given by:

v = r * w

So, w = v / r = 30 m/s / 300 m = 0.1 rad/s.

Next, we need to find the angular acceleration. The relationship between linear acceleration (a) and angular acceleration (alpha) is given by:

a = r * alpha

So, alpha = a / r = 4 m/s^2 / 300 m = 0.0133 rad/s^2.

Finally, we can substitute these values into the formula for resultant acceleration:

a = sqrt((300 m * (0.1 rad/s)^2)^2 + (0.0133 rad/s^2)^2) a = sqrt((9 m^2/s^2)^2 + (0.0133 m/s^2)^2) a = sqrt(81 m^2/s^4 + 0.000177 m^2/s^4) a = sqrt(81.000177 m^2/s^4) a = 9.00001 m/s^2

So, the magnitude of the resultant acceleration at that instant is approximately 9 m/s^2.