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Chapter 3: Problem 349

A car travelling at a speed of \(30 \mathrm{~km} / \mathrm{h}\) is brought to a halt in 8 meters by applying brakes. If the same car is travelling at $60 \mathrm{~km} / \mathrm{h}$ it can be brought to a halt with the same breaking power in (A) \(8 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(24 \mathrm{~m}\) (D) \(32 \mathrm{~m}\)

Short Answer

Expert verified
The car traveling at 60 km/h can be brought to a halt with the same braking power in \(32 \mathrm{~m}\).

Step by step solution

01

Convert speeds to m/s

To work with speeds in meters and seconds, we first need to convert the given speeds. Speed can be converted from km/h to m/s by multiplying it with 5/18: Speed at 30 km/h = \(30 * \frac{5}{18}\) m/s Speed at 60 km/h = \(60 * \frac{5}{18}\) m/s
02

Calculate deceleration

We can use the formula v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance traveled. We know that v = 0 (since the car comes to a halt), u = 30 * 5/18 m/s, and s = 8 meters. Plugging these values into the formula, we can find the deceleration: \(0 = (30 * \frac{5}{18})^2 + 2a(8)\)
03

Solve for deceleration

Now solve for "a": a = \(-\frac{(30 * \frac{5}{18})^2}{2*8}\)
04

Use deceleration to find the stopping distance at 60 km/h

Now that we have the deceleration caused by braking, we can use the same formula (v^2 = u^2 + 2as) to find the stopping distance for the car traveling at 60 km/h: 0 = \((60 * \frac{5}{18})^2 + 2 \cdot a \cdot s\) where "a" is the same deceleration found in step 3, and "s" is the distance we are looking for.
05

Solve for stopping distance

Rearrange the equation to solve for "s": s = \(-\frac{(60 * \frac{5}{18})^2}{2a}\) By substituting the value of "a" from step 3, we can find the stopping distance "s": s = 32 m So the answer is (D) \(32 \mathrm{~m}\).

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Most popular questions from this chapter

The velocity of a body of mass \(20 \mathrm{~kg}\) decrease from $20 \mathrm{~ms}^{-1}\( to \)5 \mathrm{~ms}^{-1}\( in a distance of \)100 \mathrm{~m}$. Force on the body is (A) \(-27.5 \mathrm{~N}\) (B) \(-47.5 \mathrm{~N}\) (C) \(-37.5 \mathrm{~N}\) (D) \(-67.5 \mathrm{~N}\)

A cricket ball of mass \(150 \mathrm{~g}\). is moving with a velocity of $12 \mathrm{~m} / \mathrm{s}$ and is hit by a bat so that the ball is turned back with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). If duration of contact between the ball and the bat is \(0.01 \mathrm{sec}\). The impulse of the force is (A) \(7.4 \mathrm{NS}\) (B) \(4.8 \mathrm{NS}\) (C) \(1.2 \mathrm{NS}\) (D) \(4.7 \mathrm{NS}\)

A balloon has a mass of \(10 \mathrm{~g}\) in air, The air escapes from the balloon at a uniform rate with a velocity of \(5 \mathrm{~cm} / \mathrm{s}\) and the balloon shrinks completely in \(2.5 \mathrm{sec}\). calculate the average force acting on the balloon. (A) 20 dyne (B) 5 dyne (C) 0 dyne (D) 10 dyne

Formula for true force is (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=[\\{\mathrm{d}(\mathrm{mv})\\} / \mathrm{dt}]\) (C) \(\mathrm{F}=\mathrm{m}(\mathrm{dv} / \mathrm{dt})\) (D) \(F=m\left(d^{2} x / d t^{2}\right)\)

Same forces act on two bodies of different mass \(2 \mathrm{~kg}\) and $5 \mathrm{~kg}$ initially at rest. The ratio of times required to acquire same final velocity is (A) \(5: 3\) (B) \(25: 4\) (C) \(4: 25\) (D) \(2: 5\)
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