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Truck on circular road, speed and acceleration

Short answer

Integrate a_t = 0.4t: v = 4 + 0.2t². At t=4 s: v = 7.2 m/s; a_t = 1.6 m/s²; a_n = v²/ρ = 7.2²/50 = 1.04 m/s²; a = √(1.6²+1.04²) = 1.91 m/s².

The truck travels along a circular road that has a radius of 50 m at a speed of 4 m/s. For a short distance when t=0t = 0, its speed is then increased by at=(0.4t)a_t = (0.4t) m/s2^2, where tt is in seconds. Determine the speed and the magnitude of the truck's acceleration when t=4t = 4 s.

On a circular path the acceleration has a tangential part at=v˙a_t=\dot v and a normal part an=v2/ρa_n=v^2/\rho.

Speed at t=4t=4 s

v=v0+∫040.4t dt=4+0.2t2=4+0.2(4)2=7.2 m/sv=v_0+\int_0^{4}0.4t\,dt=4+0.2t^{2}=4+0.2(4)^{2}=7.2\ \text{m/s}

Acceleration at t=4t=4 s

at=0.4(4)=1.6 m/s2an=v2ρ=7.2250=1.0368 m/s2a=at2+an2=1.62+1.03682=1.907 m/s2\begin{aligned} a_t&=0.4(4)=1.6\ \text{m/s}^2\\ a_n&=\frac{v^{2}}{\rho}=\frac{7.2^{2}}{50}=1.0368\ \text{m/s}^2\\ a&=\sqrt{a_t^{2}+a_n^{2}}=\sqrt{1.6^{2}+1.0368^{2}}=1.907\ \text{m/s}^2 \end{aligned}

Answer: v=7.2v=7.2 m/s; a=1.91a=1.91 m/s2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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