Question

speed - time graph word problems

1. a cars speed increases uniformly from 0 m/s to 20 m/s in 10 seconds.

(a) find the acceleration.
(b) how far does the car travel in this time?

1. a bus travels at a constant speed of 15 m/s for 20 s, then decelerates uniformly to rest in 10 s.

(a) what is the deceleration?
(b) use the graph to solve.
(c) find the total distance travelled.

1. a cyclist accelerates uniformly from rest to 12 m/s in 6 s, continues at this speed for 4 s, and then slows down to rest in 8 s.

(a) find the acceleration.
(b) find the deceleration.
(c) find the total distance travelled.

Explanation:

1. Car problem

Step1: Calculate acceleration

Acceleration formula is $a=\frac{v - u}{t}$, where $v$ is final - velocity, $u$ is initial - velocity and $t$ is time. Here, $u = 0$ m/s, $v = 20$ m/s, $t = 10$ s.
$a=\frac{20 - 0}{10}=2$ m/s²

Step2: Calculate distance

We can use the formula $s=ut+\frac{1}{2}at^{2}$. Since $u = 0$ m/s, the formula simplifies to $s=\frac{1}{2}at^{2}$. Substituting $a = 2$ m/s² and $t = 10$ s, we get $s=\frac{1}{2}\times2\times10^{2}=100$ m.

Step1: Calculate deceleration

Initial velocity $u = 15$ m/s, final velocity $v = 0$ m/s, time $t = 10$ s. Using the acceleration formula $a=\frac{v - u}{t}$, we have $a=\frac{0 - 15}{10}=- 1.5$ m/s².

Step2: Analyze graph for distance

The motion has two parts: constant - speed and deceleration.
For the constant - speed part: distance $s_1=v\times t_1$, where $v = 15$ m/s and $t_1 = 20$ s, so $s_1=15\times20 = 300$ m.
For the deceleration part: we can use the formula $s=ut+\frac{1}{2}at^{2}$ or the area of the triangle formed in the graph. Using the area of the triangle formula $s_2=\frac{1}{2}\times base\times height$. Here, base $t_2 = 10$ s and height $v = 15$ m, so $s_2=\frac{1}{2}\times10\times15 = 75$ m.
The total distance $s=s_1 + s_2=300+75 = 375$ m.

Step1: Calculate acceleration

Initial velocity $u = 0$ m/s, final velocity $v = 12$ m/s, time $t = 6$ s. Using $a=\frac{v - u}{t}$, we get $a=\frac{12 - 0}{6}=2$ m/s².

Step2: Calculate deceleration

Initial velocity $u = 12$ m/s, final velocity $v = 0$ m/s, time $t = 8$ s. Using $a=\frac{v - u}{t}$, we get $a=\frac{0 - 12}{8}=-1.5$ m/s².

Step3: Calculate total distance

For the acceleration part: Using $s_1=ut+\frac{1}{2}at^{2}$ with $u = 0$ m/s, $a = 2$ m/s² and $t = 6$ s, $s_1=\frac{1}{2}\times2\times6^{2}=36$ m.
For the constant - speed part: $s_2=v\times t_2$, where $v = 12$ m/s and $t_2 = 4$ s, so $s_2=12\times4 = 48$ m.
For the deceleration part: Using the formula $s=ut+\frac{1}{2}at^{2}$ or the area of the triangle. Using the area of the triangle formula with base $t_3 = 8$ s and height $v = 12$ m, $s_3=\frac{1}{2}\times8\times12 = 48$ m.
The total distance $s=s_1 + s_2+s_3=36 + 48+48=132$ m.

Answer:

(a) $2$ m/s²
(b) $100$ m

2. Bus problem