Question:
A car is moving along a straight line and is brought to a stop within a distance of \(200\) m and in time \(10\) s. The initial speed of the car is
A car is moving along a straight line and is brought to a stop within a distance of \(200\) m and in time \(10\) s. The initial speed of the car is
Show Hint
When a body comes to rest under uniform deceleration, the average velocity is
\[
\frac{u+v}{2}.
\]
Using
\[
s=\frac{u+v}{2}\,t
\]
often provides the quickest solution.
Updated On: Jun 26, 2026
- \(25\ \text{m s}^{-1}\)
- \(50\ \text{m s}^{-1}\)
- \(75\ \text{m s}^{-1}\)
- \(40\ \text{m s}^{-1}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Write the given data.
The car is brought to rest, therefore the final velocity is \[ v=0. \] The distance covered before stopping is \[ s=200\ \text{m}. \] The time taken is \[ t=10\ \text{s}. \] Let the initial velocity be \[ u. \]
Step 2: Use the equation of motion.
For uniformly accelerated motion, \[ s=\frac{(u+v)}{2}\,t. \] Substituting the given values, \[ 200=\frac{(u+0)}{2}\times 10. \] \[ 200=5u. \]
Step 3: Calculate the initial velocity.
\[ u=\frac{200}{5}. \] \[ u=40\ \text{m s}^{-1}. \]
Step 4: Verification using acceleration.
Using \[ v=u+at, \] we get \[ 0=40+10a. \] \[ a=-4\ \text{m s}^{-2}. \] Now, \[ s=ut+\frac{1}{2}at^2 \] \[ =40(10)+\frac{1}{2}(-4)(10)^2 \] \[ =400-200 \] \[ =200\ \text{m}. \] Hence, the result is verified.
Step 5: Final conclusion.
Therefore, the initial speed of the car is \[ \boxed{40\ \text{m s}^{-1}} \] Hence, the correct option is \[ \boxed{(4)} \]
The car is brought to rest, therefore the final velocity is \[ v=0. \] The distance covered before stopping is \[ s=200\ \text{m}. \] The time taken is \[ t=10\ \text{s}. \] Let the initial velocity be \[ u. \]
Step 2: Use the equation of motion.
For uniformly accelerated motion, \[ s=\frac{(u+v)}{2}\,t. \] Substituting the given values, \[ 200=\frac{(u+0)}{2}\times 10. \] \[ 200=5u. \]
Step 3: Calculate the initial velocity.
\[ u=\frac{200}{5}. \] \[ u=40\ \text{m s}^{-1}. \]
Step 4: Verification using acceleration.
Using \[ v=u+at, \] we get \[ 0=40+10a. \] \[ a=-4\ \text{m s}^{-2}. \] Now, \[ s=ut+\frac{1}{2}at^2 \] \[ =40(10)+\frac{1}{2}(-4)(10)^2 \] \[ =400-200 \] \[ =200\ \text{m}. \] Hence, the result is verified.
Step 5: Final conclusion.
Therefore, the initial speed of the car is \[ \boxed{40\ \text{m s}^{-1}} \] Hence, the correct option is \[ \boxed{(4)} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Mechanics Questions
- A body of mass 5 kg is moving with a velocity of 10 m/s. What is its kinetic energy?
- A 200 kg steel bar sits horizontally on two supports as shown in the figure. The force on each support is (\( g = 9.8 \, \text{ms}^{-2} \))
- A sphere rolls down from the top of an inclined plane which makes an angle $30^\circ$ with the edge of a horizontal roof of a house. If the highest and lowest points of the inclined plane are 8.75 m and 3.75 m respectively from the ground then the horizontal distance from the lower edge of the roof at which the sphere hits the ground is (acceleration due to gravity = 10 m s$^{-2}$)
- At the moment $t = 0$, a time dependent force $F = at$ (where $a$ is constant equal to 1 N s$^{-1}$) is applied on a body of mass 1 kg resting on a smooth horizontal plane as shown in the figure. If the direction of this force makes an angle $45^\circ$ with the horizontal, then the velocity of the body at the moment it leaves the plane is (acceleration due to gravity = 10 m s$^{-2}$)
- An object of mass 3 kg is tied by a string of negligible mass to a ceiling and held such that the string is taut. The object is released suddenly such that the string remains taut. It’s acceleration when released is (acceleration due to gravity = 10 m s$^{-2}$)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions