Question

A car accelerates with a constant acceleration of 1.8 \text{ m}\cdot\text{s}^{-2}1.8 m⋅s
−2
along a long straight road. How long does it take the car to accelerate from 100 kph to 110 kph, and how far does it travel while doing so? [Hint: watch your units and sig figs! Because there is no other information, take the speeds as three significant figures]

It takes ______ seconds to accelerate to 110 kph. While accelerating, the car travels _____ m.

Answer 1

Given that,

Acceleration of the car, $a=1.8\ m/s^2$

Initial speed of the car, u = 100 km/h = 27.77 m/s

Final speed of the car, v = 110 km/h = 30.55 m/s

Solution,

Let t is the time taken by the car to accelerate from u to v. Using formula of acceleration as :

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{30.55-27.77}{1.8}\\\\t=1.55\ s$

Let d is the distance covered by the car bu doing so. Using third equation of kinematics as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(30.55)^{2}-(27.77)^{2}}{2\times1.8}\\\\d=45.03\ m$

Learn more,

Kinematics

https://brainly.in/question/16576401

Answer 2

Answer:

Answer ⇒ Option (c). The minimum stopping distance is same for both the cars.

Explanation ⇒  Let us take an help of the mathematical expression to derive our result.

We know that when car will be stopped then the Force of Friction acting will be the Limiting Friction.

∴ f = μmg

Now, ma =  μmg

⇒ a =  μg

Since, the acceleration is independent of the mass thus the minimum stopping distance will also be independent of the mass as they have the same initial speed. [Taking help of the Second Equation of motion.]

∴ Option (c). is correct answer.