Explain the distinction between speed and velocity - Leaving Cert Physics - Question a - 2014

To find the time taken to reach the top speed, we use the formula:

$t = \frac{(v - u)}{a}$

Where:

• Final speed, $v = 15 \, \text{m/s}$
• Initial speed, $u = 0 \, \text{m/s}$
• Acceleration, $a = 0.5 \, \text{m/s}^2$

Substituting the values:

$t = \frac{(15 - 0)}{0.5} = 30 \, \text{seconds}$

The distance traveled at top speed can be calculated using the formula:

$d = v \times t$

Where:

• Speed, $v = 15 \, \text{m/s}$
• Time, $t = 100 \, \text{seconds}$

Thus:

$d = 15 \times 100 = 1500 \, \text{meters}$

To find the acceleration required to stop the bus, we can use:

$a = \frac{(v - u)}{t}$

Where:

• Final speed, $v = 0 \, \text{m/s}$ (since the bus comes to a stop)
• Initial speed, $u = 15 \, \text{m/s}$
• Time taken to stop, $t = 20 \, \text{seconds}$

Substituting the values:

$a = \frac{(0 - 15)}{20} = -0.75 \, \text{m/s}^2$

(The negative sign indicates that the bus is decelerating.)

The velocity-time graph would be a piecewise linear graph:

• The graph rises linearly from the origin to $15 \, \text{m/s}$ over $30 \, \text{seconds}$ (acceleration phase).
• It remains constant at $15 \, \text{m/s}$ for $100 \, \text{seconds}$ (constant speed phase).
• The graph then slants down to $0 \, \text{m/s}$ over $20 \, \text{seconds}$ (deceleration phase).

The segments will represent an increasing line, a flat constant line, and a decreasing line, showing the different phases of the bus's motion.