Question

A car is stationary. It accelerates at 0.8 ms^2 for 10 s and then at 0.4 ms^2 for a further 10 s. Use the equations of motion to deduce the car's final displacement. You will have to split the journey into two parts, since the acceleration changes after 10 s.

Answer 1

Answer:

The final displacement of the car is 140 meters.

Explanation:

The final displacement of the car ($s$), in meters, is the sum of the change in displacement associated with each part of the journey, which is derived from the following kinematic formulas:

$s = s_{1} + s_{2}$ (1)

$s_{1} = \frac{1}{2}\cdot a_{1}\cdot t_{1}^{2}$ (2)

$v_{o,2} = a_{1}\cdot t_{1}$ (3)

$s_{2} = v_{o,2} + \frac{1}{2}\cdot a_{2}\cdot t_{2}^{2}$ (4)

Where:

$s_{1}$ - Traveled distance of the first part, in meters.

$s_{2}$ - Traveled distance of the second part, in meters.

$a_{1}$ - Acceleration in the first part, in meters per square second.

$a_{2}$ - Acceleration in the second part, in meters per square second.

$v_{o,2}$ - Initial speed of the car in the second part, in meters per second.

$t_{1}$ - Time taken in the first part, in seconds.

$t_{2}$ - Time taken in the second part, in seconds.

If we know that $a_{1} = 0.8\,\frac{m}{s^{2}}$, $t_{1} = 10\,s$, $a_{2} = 0.4\,\frac{m}{s^{2}}$ and $t_{2} = 10\,s$, then the distance traveled by the car is:

By (2):

$s_{1} = \frac{1}{2}\cdot \left(0.8\,\frac{m}{s^{2}} \right)\cdot(10\,s)^{2}$

$s_{1} = 40\,m$

By (3):

$v_{o,2} = \left(0.8\,\frac{m}{s^{2}} \right)\cdot (10\,s)$

$v_{o,2} = 8\,\frac{m}{s}$

By (4):

$s_{2} = \left(8\,\frac{m}{s} \right)\cdot (10\,s) + \frac{1}{2}\cdot \left(0.4\,\frac{m}{s^{2}} \right)\cdot (10\,s)^{2}$

$s_{2} = 100\,m$

By (1):

$s = 140\,m$

The final displacement of the car is 140 meters.