Pizza War 1

Classical Mechanics Level 3

Vaibhav gives a challenge to Kalash. He says that describe a motion such that the velocity vector at some instant of time makes an angle of $45^{\circ}$ with the acceleration vector. Kalash then swiftly answers that $\vec{r} = at\hat{i} + at\left(1 - \dfrac{t}{10}\right)\hat{j}$ .

Kalash then asks Vaibhav at what instant of time will this occur for the motion described above. Vaibhav is confused. Can you help him out?

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2 solutions

Hem Shailabh Sahu
May 18, 2015

We know that the Velocity Vector, $\vec{v}=\frac{\mathrm d}{\mathrm dt} \left (\vec{r} \right)$ & Acceleration Vector, $\vec{A}=\frac{\mathrm d}{\mathrm dt} \left (\vec{v} \right)$

As we know, from the properties of Scalar/Dot Product of Two vectors, $\vec{v}.\vec{A}=\vec{A}.\vec{v}=|\vec{v}|.|\vec{A}|.\cos(\theta)$ where $\theta$ is the angle between $\vec{v}$ & $\vec{A}$ ; $|\vec{v}|$ is the Magnitude of the Velocity Vector, $\vec{v}$ .

Putting the Angle Between $\vec{v}$ & $\vec{A}$ as $45^{\circ}$ , we get the Instant $t$ when the condition (Angle between $\vec{v}$ & $\vec{A}$ is $45^{\circ}$ ) is followed as, $t=10 seconds$ $or$ $t=0 seconds$ Neglecting $t=0 seconds$ (as not in the options), we get $t=\boxed{10}$

The problem is overrated too!

Hem Shailabh Sahu - 6 years ago

We can do it using the cross product as well but yes,taking the dot product is much easier.

Ayon Ghosh - 3 years, 8 months ago

actually in t=0 the acceleration vector and velocity vector will make an angle of 135° not 45° so only t=10 is the answer

Sahar Bano - 5 months ago

Zebern Jethro Meguillo
Sep 8, 2015

Just scroll down to see solution

Pizza War 1

2 solutions

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