A car is moving at a speed of 40 , m/s on a c | vedclass

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(C) Given:Speed $v = 40 \, m/s$Radius $r = 400 \, m$Tangential acceleration $a_T = 3 \, m/s^2$The centripetal acceleration $a_c$ is given by the formu

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$5$

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(C) Given:Speed $v = 40 \, m/s$Radius $r = 400 \, m$Tangential acceleration $a_T = 3 \, m/s^2$The centripetal acceleration $a_c$ is given by the formula:$a_c = \frac{v^2}{r} = \frac{40^2}{400} = \frac{1600}{400} = 4 \, m/s^2$The total acceleration $a$ is the vector sum of the centripetal and tangential accelerations:$a = \sqrt{a_c^2 + a_T^2}$$a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, m/s^2$Therefore,the total acceleration of the car is $5 \, m/s^2$.

$A$ car is moving at a speed of $40 \, m/s$ on a circular track of radius $400 \, m$. This speed is increasing at the rate of $3 \, m/s^2$. The acceleration of the car is ....... $m/s^2$.

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