The kinetic energy of a car is doubled when i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the initial velocity be $v$. Initial kinetic energy $E = \frac{1}{2}mv^2$. When the velocity is increased by $1 \,m/s$, the new velocity becom

Vedclass · Accepted Answer

$(1+\sqrt{2}) \,m/s$

Vedclass · Answer

(D) Let the initial velocity be $v$. Initial kinetic energy $E = \frac{1}{2}mv^2$. When the velocity is increased by $1 \,m/s$, the new velocity becomes $v' = v + 1$. The new kinetic energy $E' = \frac{1}{2}m(v+1)^2$. According to the problem, the kinetic energy is doubled, so $E' = 2E$. Substituting the expressions: $\frac{1}{2}m(v+1)^2 = 2 	imes (\frac{1}{2}mv^2)$. Simplifying the equation: $(v+1)^2 = 2v^2$. Expanding the left side: $v^2 + 2v + 1 = 2v^2$. Rearranging the terms to form a quadratic equation: $v^2 - 2v - 1 = 0$. Using the quadratic formula $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $v = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$. Since velocity must be positive, we take $v = (1 + \sqrt{2}) \,m/s$.

The kinetic energy of a car is doubled when its velocity is increased by $1 \,m/s$. Then the initial velocity of the car is

Similar Questions

What is the shape of the graph between speed and kinetic energy of a body?

Two particles having masses $4\, g$ and $16\, g$ respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is $n : 2$. The value of $n$ will be ...... .

$A$ man increases his speed by $2 \ m/s$, and as a result, his kinetic energy doubles. What was his initial speed?

Two bodies having kinetic energy in the ratio $4: 1$ are moving with the same linear velocity. The ratio of their masses is

If the linear momentum is increased by $5\%$,the kinetic energy will increase by.......$\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module