A car moving with a speed of 72 kmph is brought to rest by applying brakes in 10 sec. Find the retardation. (JEE Main 2018)
The answer is 5 m/s^2.
The formula for retardation is:
$$a = \frac{v_f - v_i}{t}$$
In this problem, $v_f = 0$, $v_i = 72 \text{ kmph} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$, and $t = 10 \text{ sec}$. Substituting these values into the formula, we get:
$$a = \frac{0 - 20}{10} = -2 \text{ m/s}^2$$
The negative sign indicates that the car is decelerating.
**Two masses m1 and m2 are connected by a light string passing over a frictionless pulley. If the pulley is accelerating upwards with an acceleration a, the tension in the string will be
