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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line4 Marks
Question
Read the passage given below and answer the following questions from $1$ to $5$. When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below: $v = u + at $$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$ $2a s = v^2 – u^2$ Where u is the initial velocity of the object which moves with uniform acceleration a for Time $t, v$ is the final velocity and s is the distance travelled by the object in time.
  1. equation of motions are applicable to motion with
  1. uniform acceleration
  2. non uniform acceleration
  3. constant velocity
  4. none of these
  1. There are $4$ equation of motion. True or false?
  1. True
  2. False
  1. The brakes applied to a car produce an acceleration of $10\ m/s^2$​​​​​​​ in the opposite direction to the motion. If the car takes $1\ s$ to stop after the application of brakes, calculate the distance traveled during this time by car.
  1. An object is dropped from a tower falls with a constant acceleration of $10\ m/s2$. Find its speed $10\ s$ after it was dropped.
  1. A bullet hits a Sand box with a velocity of $10\ m/s$ and penetrates it up to a distance of $5\ cm$. Find the deceleration of the bullet in the sand box
✓

Answer

  1. (a) uniform acceleration
  2. (b) False
  3. Here in this problem,
$v = 0$
$a = -10 m/s2$ (as acceleration is retarding)
$t = 1\ sec.$
To find: distance travelled
Solution: using kinematic equation
$v = u + at$
$0= u + -10 \times 1$
$u = 10\ m/s$
Therefore distance is given by
$\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$
$\text{S}=10\times1-\Big(\frac{1}{2}\Big)\times10\times1^2$
$s = 5m$
  1. Here in this problem,
$u = 0$
$a = 10 m/s^2​​​​​​​$​​​​​​​(as acceleration is in the direction of gravity)
$t = 10\ sec$.
To find: final velocity after 10 second
Solution: using kinematic equation
$v = u + at$
$v = 0 + 10 \times 10$
$v = 100 m/s$​​​​​​​
  1. Here in this problem,
$v = 0$(as bullet is going to stop)
$u = 10 m/s$
$s = 5m$.
To find: deceleration of the bullet
Solution: using kinematic equation
$2a s = v^2 – u^2$
$2 \times a \times 5 = 0^2- 10^2$
$10a = -100$
$\text{A}=\frac{100}{10}$
$a = -10m/s^2$. Negative sign indicates that it is deceleration.

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