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The acceleration-time relationship for a vehicle subjected to non-uniform acceleration is, $\frac{d v}{d t}=\left(\alpha-\beta v_{0}\right) e^{-\beta t}$ where, $v$ is the speed in $\mathrm{m} / \mathrm{s}, t$ is the time in $\mathrm{s}, \alpha$ and $\beta$ are parameters, and $v_{0}$ is the initial speed in $\mathrm{m} / \mathrm{s}$. If the accelerating behavior of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as, $\frac{d v}{d t}=(\alpha-\beta v)$ Considering $\alpha=2 \mathrm{~m} / \mathrm{s}^{2}, \beta=0.05 \mathrm{~s}^{-1}$ and $\frac{d v}{d t}=1.3 \mathrm{~m} / \mathrm{s}^{2}$ at $t=3 \mathrm{~s}$, the distance (in $\mathrm{m}$ ) travelled by the vehicle in $35 \mathrm{~s}$ is


Question ID - 156136 | Toppr Answer

The acceleration-time relationship for a vehicle subjected to non-uniform acceleration is, $\frac{d v}{d t}=\left(\alpha-\beta v_{0}\right) e^{-\beta t}$ where, $v$ is the speed in $\mathrm{m} / \mathrm{s}, t$ is the time in $\mathrm{s}, \alpha$ and $\beta$ are parameters, and $v_{0}$ is the initial speed in $\mathrm{m} / \mathrm{s}$. If the accelerating behavior of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as, $\frac{d v}{d t}=(\alpha-\beta v)$ Considering $\alpha=2 \mathrm{~m} / \mathrm{s}^{2}, \beta=0.05 \mathrm{~s}^{-1}$ and $\frac{d v}{d t}=1.3 \mathrm{~m} / \mathrm{s}^{2}$ at $t=3 \mathrm{~s}$, the distance (in $\mathrm{m}$ ) travelled by the vehicle in $35 \mathrm{~s}$ is

1 Answer - 5876 Votes

3537

Answer Key : (895 to 905) -

895 to 905



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