Question

For the given velocity-time graph the displacement
of the object between t = 2 s to t = 4 s is

(1) 6 m
(2) 10 m
(4) 2 m
(3) 18 m​

Answer 1

Answer:

An object is thrown vertically upwards and

rises to a height of 10 m. Calculate

i. the velocity with which the object was thrown

upwards and

ii. the time taken by the object to reach the highest

point. (g = 9.8 m/s)

An object is thrown vertically upwards and

rises to a height of 10 m. Calculate

i. the velocity with which the object was thrown

upwards and

ii. the time taken by the object to reach the highest

point. (g = 9.8 m/s)

$solve$

Answer 2

Answer:

For the given velocity-time graph the displacement of the object between t = 2 s to t = 4 s is 6m.

Option (1) is the correct answer.

Explanation:

Given, Velocity - time graph with velocity on Y-axis and time on X- axis. Area under the velocity - time graph represents the displacement.

In order to find the displacement, that is area under the curve, we should first determine the equation of the line in the graph.

• Since two points on the curve are known (0,0) and (6,6).

                The slope of the line = $\frac{6-0}{6-0}$ = 1

• Since the line is passing through the origin, the equation of line is of the form y = mx, where m is slope of the line

                Therefore, the equation of line is v = t.

To find the displacement from t = 2s to t = 4s, we should integrate the line equation with respect to time.

Displacement = $\int\limits^4_2 {v} \, dt$  

                       = $\int\limits^4_2 {t} \, dt$

                       = $\frac{t^{2} }{2} \left \{ {{t=4} \atop {t=2}} \right.$

                       = (16-4)/2

Displacement = 6 m

Therefore, the displacement of the object between t = 2 s to t = 4 s is 6 m.

Option (1) is the correct answer.

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