A truck is moving at a speed of 54 km/h. The | vedclass

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(A) Given: Initial velocity $v_1 = 54 \ km/h = 54 	imes \frac{5}{18} \ m/s = 15 \ m/s$. Radius $r = 50 \ cm = 0.5 \ m$. Number of rotations $n = 20$.

Vedclass · Accepted Answer

$62.8 \ m, -3.58 \ rad \ s^{-2}$

Vedclass · Answer

(A) Given: Initial velocity $v_1 = 54 \ km/h = 54 	imes \frac{5}{18} \ m/s = 15 \ m/s$. Radius $r = 50 \ cm = 0.5 \ m$. Number of rotations $n = 20$. Total angular displacement $	heta = 20 	imes 2\pi \ rad = 40\pi \ rad$. Initial angular velocity $\omega_1 = \frac{v_1}{r} = \frac{15}{0.5} = 30 \ rad \ s^{-1}$. Final angular velocity $\omega_2 = 0$. Using the equation $\omega_2^2 = \omega_1^2 + 2\alpha	heta$: $0 = (30)^2 + 2 	imes \alpha 	imes 40\pi$ $\alpha = -\frac{900}{80\pi} \approx -3.58 \ rad \ s^{-2}$. Linear distance $d = n 	imes (2\pi r) = 20 	imes 2 	imes 3.14 	imes 0.5 = 62.8 \ m$.

$A$ truck is moving at a speed of $54 \ km/h$. The radius of its wheels is $50 \ cm$. If the wheels come to rest after $20$ rotations upon applying brakes,what is the linear distance covered by the truck during this time? What is the angular acceleration of the wheels?

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