Question

1. The position of a particle moving in a straight line along x-axis varies with time t as x = (12t - 3t²) m, where to is in second. The distance covered by particle in first 5 s is :-

2. Two cars are moving towards each other on a straight road with acceleration 5 m/ s². At a certain instant, in their separation is 400 m and their velocities are 10 m/s and 20 m/s. How long will it take for the cars to collide?

Answer 1

Answer:

please refer to the attachment for answers.

Answer 2

The distance covered by particle in first 5 s is 3.6m and the time taken by the cars to collide is 5s.

Given:

For the first question, the given details are as follows:

x = (12t - 3t²) m.

Time, t = 5s.

For the second question, the given details are as follows:

The acceleration of first car, $a_1$ = 5 m/ s².

The acceleration of second car, $a_2$ = 15 m/ s².

The distance separation between two cars = 400 m.

The velocity of first car, $u_1$ = 10 m/s.

The velocity of second car, $u_2$ = 20 m/s.

To Find:

We have to find the distance covered by particle in first 5 for the first question and the time taken by the cars to collide for the second question.

Solution:

1. Consider the first question.

Given that, x = (12t - 3t²) m.

Or, x = t × (12 - 3t)

t × (12 - 3t) = 0 at t = 4 or t = 0.

On differentiating x, we get,

$\frac{dx}{dt} = 12 - 2 . 3t = v.$

The distance covered by the particle, d = $\frac{Velocity }{t}$

At t = 5,

$d = \frac{12 - 6t}{t} = \frac{12 - (6).(5)}{5} = \frac{-18}{5} = -3.6$

Since distance cannot be negative, we can take it as positive.

∴, $d = 3.6m$.

2. Consider the second question.

Suppose the two cars collide after t seconds. Then they will be at the same point after the time t.

Let $d_1$ be the distance of first car and $d_2$ be the distance of second car.

∴, $d_1 + d_2 = 400$

Using the equation of motion $ut + \frac{1}{2} at^2$, the above equation can be rewritten as,

$(u_1t + \frac{1}{2} a_1t^2) + (u_2t + \frac{1}{2} a_2t^2) = 400$

Substituting the given values in the above equation, we get,

$10t + \frac{1}{2} 5t^2 + 20t + \frac{1}{2} 15t^2 = 400$

On simplifying, the above equation becomes,

$30t + \frac{20}{2} t^2 = 400$

Or, $30t + 10 t^2 = 400$

$t^2 + 3t - 4 = 0$

i.e., $(t + 8) . (t + 5) = 0$

Thus we get, t = -8 or t = 5.

Time cannot be negative.

∴, Time of collapse, t = 5s.

Hence, the correct answers are:

The distance covered by particle in first 5 s = 3.6m.

The time taken by the cars to collide = 5s.

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