Question

A car is moving along a straight line is brought to a stop within a distance of 200 m and in time 10 s. The initial speed of the car is

• A. \( 25 \text{ ms}^{-1} \)
• B. \( 50 \text{ ms}^{-1} \)
• C. \( 75 \text{ ms}^{-1} \)
• D. \( 40 \text{ ms}^{-1} \)

Hint:

Equations of motion for constant acceleration:
\(v = u + at\)
\(s = ut + \frac{1}{2}at^2\)
\(v^2 = u^2 + 2as\)
\(s = \frac{u+v}{2}t\)
Identify knowns and unknowns, then choose the appropriate equation.
"Brought to a stop" means final velocity \(v=0\).

Answer 1

The Correct Option is D

Let the initial speed of the car be \(u\). The car is brought to a stop, so its final speed \(v = 0\). Distance covered \(s = 200\) m. Time taken \(t = 10\) s. We can use the equation of motion: \(s = \left(\frac{u+v}{2}\right)t\), which is valid for constant acceleration (deceleration here). Substitute the given values: \(200 = \left(\frac{u+0}{2}\right) \times 10\) \(200 = \frac{u}{2} \times 10\) \(200 = 5u\) \(u = \frac{200}{5} = 40 \text{ ms}^{-1}\). The initial speed of the car is \(40 \text{ ms}^{-1}\). This matches option (d). We can also find the acceleration (deceleration) \(a\): Using \(v = u+at \Rightarrow 0 = u + a(10) \Rightarrow u = -10a\). Using \(s = ut + \frac{1}{2}at^2 \Rightarrow 200 = u(10) + \frac{1}{2}a(10)^2 = 10u + 50a\). Substitute \(u=-10a\): \(200 = 10(-10a) + 50a = -100a + 50a = -50a\). \(a = \frac{200}{-50} = -4 \text{ ms}^{-2}\) (deceleration). Then \(u = -10a = -10(-4) = 40 \text{ ms}^{-1}\). \[ \boxed{40 \text{ ms}^{-1}} \]