Question

The speed time graph shows the motion of toboggan as it descends a hill. Determine the acceleration of the toboggan between,

a) a and b
b) b and c
c) c and d

Answer 1

B (B AND C)

Explanation:

BECAUSE THE POINT B IS HIGH ND I SUPPOSE THAT IT IS SPEED.

Answer 2

The acceleration of the toboggan between:

a) points a and b is 8.33 ft/s^2

b) points b and c is -5 ft/s^2

c) points c and d is 0 ft/s^2

The acceleration of the toboggan can be calculated by finding the slope of the speed-time graph between the given points.

a) The acceleration between points A and B can be determined by calculating the change in velocity over the change in time.

`\text{Acceleration}_(AB) = \frac{\text{Change in Velocity}_(AB)}{\text{Change in Time}_(AB)} \]\\\text{Acceleration}_(AB) = \frac{(35 \, \text{m/s} - 10 \, \text{m/s})}{(3 \, \text{s} - 0 \, \text{s})} \]\\\text{Acceleration}_(AB) = \frac{25 \, \text{m/s}}{3 \, \text{s}} \]\\\text{Acceleration}_(AB) \approx 8.33 \, \text{m/s}^2 \]`

b) The acceleration between points B and C can be calculated similarly:

`\[ \text{Acceleration}_(BC) = \frac{\text{Change in Velocity}_(BC)}{\text{Change in Time}_(BC)} \]\\ \text{Acceleration}_(BC) = \frac{(20 \, \text{m/s} - 35 \, \text{m/s})}{(6 \, \text{s} - 3 \, \text{s})} \]\\\text{Acceleration}_(BC) = \frac{-15 \, \text{m/s}}{3 \, \text{s}} \]\\\text{Acceleration}_(BC) \approx -5 \, \text{m/s}^2 \]`

c) The acceleration between points C and D is calculated the same way:

`\[ \text{Acceleration}_(CD) = \frac{\text{Change in Velocity}_(CD)}{\text{Change in Time}_(CD)} \]\\\text{Acceleration}_(CD) = \frac{(20 \, \text{m/s} - 20 \, \text{m/s})}{(6 \, \text{s} - 6 \, \text{s})} \]\\ \text{Acceleration}_(CD) = \frac{0 \, \text{m/s}}{0 \, \text{s}} \]`

Therefore the acceleration of the toboggan between:

a) points a and b is 8.33 ft/s^2

b) points b and c is -5 ft/s^2

c) points c and d is 0 ft/s^2