Problem

The truck travels along a circular road that has a radius of 50 m at a speed of 4 m/s. F...

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 = 0, its speed is then increased by where t is in seconds. Determine the speed and the magnitude of the truck's acceleration when t = 4 s.

Step-by-Step Solution

Solution 1

Given

Radius of the path, \(\rho=50 \mathrm{~m}\)

Initial speed, \(v_{1}=4 \mathrm{~m} / \mathrm{s}\)

Acceleration, \(a_{t}=(0.4 t) \mathrm{m} / \mathrm{s}^{2}\)

$$ \begin{aligned} &a_{t}=0.4 t \\ &\frac{d v}{d t}=0.4 t \Rightarrow d v=0.4 t d t \\ &\int_{4}^{v} d v=0.4 \int_{0}^{t} t d t \\ &v-4=0.2 t^{2} \\ &v=4+0.2 t^{2} \end{aligned} $$

when \(t=4 \mathrm{~s}\)

$$ v=4+0.2(4)^{2}=7.2 \mathrm{~m} / \mathrm{s} $$

Normal aceleration, \(a_{n}=\frac{v^{2}}{\rho}=\frac{7.2^{2}}{50}=1.0368 \mathrm{~m} / \mathrm{s}^{2}\)

Tangential aceleration, \(a_{4}=(0.4 \times 4)=1.6 \mathrm{~m} / \mathrm{s}^{2}\)

Aceleration, \(a=\sqrt{a_{n}{ }^{2}+a_{t}{ }^{2}}=\sqrt{1.0368^{2}+1.6^{2}}=1.9065 \mathrm{~m} / \mathrm{s}^{2}\)

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