Question

Ans. 100 m, - 2 m/s.
The time-velocity graph of a car is shown in the figure.
Calculate : (i) acceleration from A to B, (ii) distance
travelled in the last 4 seconds.
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Answer 1

☆ Velocity-time graph.

》Slope of velocity-time graph gives acceleration.

》Area of velocity-time graph gives displacement covered.

☆Solution:

(i) acceleration from A to B

• $\ slope = tan\theta = \dfrac{dy}{dx}$

• From A to B,

1. change in y(dy) = 5–2 = 3
2. change in x(dx) = 3–2 = 1

•$\ slope = \dfrac{3}{1} = 3$

☆Slope is always measured from positve x-axis.

•$\bf slope = \alpha = -3$

• acceleration = – 3 m/s²

(ii) distance travelled in the last 4 seconds.

• last 4 secs will be from 4s to 7s.

•Area of that region is of tapezium.

• distance = area of trapezium.

• $\rm distance = \dfrac{1}{2}×(sum\:of\:parallel\:sides)×height$

•$\rm distance = \dfrac{1}{2}(1+3)(2)$

•$\bf distance = 4m$

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