Average velocities for time intervals t = 0 to 2 s, t = 2 to 4 s and t = 0 to 4 s are respectively equal to

2.5 m/s, -2.5 m/s, 2.5 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.25 m/s, -1.25 m/s, 0 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 m/s, -5 m/s, 0 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5 m/s, -2.5 m/s, 0 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C 2.5 m/s, -2.5 m/s, 0 m/s
Displacement in a time interval = Area enclosed in the time interval
Average velocity = Total Displacement / total time

t = 0s to t = 2s:
Displacement $= \frac{1}{2} \times 2 \times 5 = 5 m$
Average velocity $= 5 / 2 = 2.5 m / s$
t = 2s to t = 4s:
Displacement $= \frac{1}{2} \times 2 \times - 5 = - 5 m$
Average velocity $= - 5 / 2 = - 2.5 m / s$
t = 0s to t = 4s:
Displacement $= 5 m - 5 m = 0 m$
Average velocity $= 0 / 4 = 0 m / s$

Suggest Corrections

Similar questions

\frac{x^{2}}{25} - 4 = 0

Q. The line

3 x + 5 y = 15 \sqrt{2}

is a tangent to the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1

at a point whose eccentric angle is

Q. If the two lines

(a + b) x - 4 y + 5 = 0

and

x + (a - b) y + 10 = 0

are perpendicular to each other then

Q. The position versus time graph for a certain particle moving along the x-axis is shown in the figure. Find the average velocity in the time intervals (a) 0 to 2 s, (b) 2 s to 4 s, and (c) 4 s to 7 s.

A coil having an area $A_{0}$ is placed in a magnetic field which changes from $B_{0}$ to $3 B_{0}$ in a time interval t. The e.m.f. induced in the coil will be

Join BYJU'S Learning Program