The velocity – time graph of a particle moving along a straight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to 5ms−2. If the average velocity during the motion is 20ms−1, then
The value of t is
A
5s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
10s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
20s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5√2s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A5s Slope of v−t graph = acceleration
From t=0 to t=t sec
Slope of line segment = 5 ⇒vt=5⇒v=5t
Since the velocity is always positive, we can say Displacement = Distance covered ⇒Average velocity = Average speed ∴ Average speed = 20ms−1
⇒Total distanceTotal time=20 ⇒Area of graph20+t=20
Area under the graph =12×5t×t+5t×(20−t)+12×5t×t=100t ⇒100t=20(20+t) ⇒t=5s