Question

Calculate the a) deceleration caused by the body
b) total distance travelled in 15 second. ​

Answer 1

Answer:

b) the total distance is 60 r

Explanation:

may it help u

Answer 2

According to this velocity-time graph, we are asked to calculate the following

a) Deceleration caused by the body

Explanation: In the velocity-time graph the slope tell us about the change in velocity per time. If the slope is increasing then it tell us about positive acceleration and if the slope is decreasing then it tell us about negative acceleration that is the retardation or the deceleration.

Here, the final velocity is 0 metre per second, the initial velocity is 4 metre per second and the time taken is 15 seconds.

Required solution:

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-4}{15} \\ \\ :\implies \sf a \: = \dfrac{-4}{15} \\ \\ :\implies \sf a \: = -0.26 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -0.26 \: ms^{-2} \\ \\ :\implies \sf Deceleration \: = -0.26 \: ms^{-2}$

________________________________

b) Total distance travelled in 15 second

Explanation: In the velocity-time graph the area under the curve tell us about the distance or the displacement. So we have to find out the area of the figure to get the distance travelled.

Knowledge required:

• Given figure is of triangle△
• Area of△is ½ × B × H

Where, B denotes base and H denotes height of the triangle.

Required solution:

$:\implies \sf Distance \: = Area \: inclosed \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times Base \times Height \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times (15-0) \times (4-0) \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 15 \times 4 \\ \\ :\implies \sf Distance \: = \dfrac{1}{\cancel{{2}}} \times 15 \times \cancel{4} \\ \\ :\implies \sf Distance \: = 15 \times 2 \\ \\ :\implies \sf Distance \: = 30 \: metres$