Question

A 2.7x10^3 kg car accelerates from rest under the action of two forces. One is a forward force of 1154 N provided by traction between the wheels and the road. The other is a 915 N resistive force due to various frictional forces. How far must the car travel for its speed to reach 2.1 m/s? Answer in units of m

Answer 1

Answer on Question#37800 - Physics - Mechanics

A $2.7 \times 10^{3} \mathrm{~kg}$ car accelerates from rest under the action of two forces. One is a forward force of $1154 \mathrm{~N}$ provided by traction between the wheels and the road. The other is a $915 \mathrm{~N}$ resistive force due to various frictional forces. How far must the car travel for its speed to reach $2.1 \mathrm{~m} / \mathrm{s}$? Answer in units of m

Solution:

Net force acting on Car:

$$F_{\text{net}} = 1154 \mathrm{~N} - 915 \mathrm{~N} = 239 \mathrm{~N}$$

So acceleration (from the Newton's second law) a is given by:

$$a = \frac{F_{\text{net}}}{m} = \frac{239 \mathrm{~N}}{2.7 \times 10^{3} \mathrm{~kg}} = 0.89 \frac{\mathrm{m}}{\mathrm{s}^{2}}$$

Rate equation for car:

$$V = a t \Rightarrow t = \frac{V}{a}$$

Equation of motion for the car:

$$S = \frac{a t^{2}}{2} = \frac{a}{2} \cdot \left(\frac{V}{a}\right)^{2} = \frac{V^{2}}{2a} = \frac{\left(2.1 \frac{\mathrm{m}}{\mathrm{s}}\right)^{2}}{2 \cdot 0.89 \frac{\mathrm{m}}{\mathrm{s}^{2}}} = 2.5 \mathrm{~m}$$

Answer: the car traveled distance $2.5\mathrm{m}$