The following graph represents the change in velocity
(magnitude of velocity) of a bus with respect to time:
a. Calculate the total distance covered by the bus in 0 s-10 s.
b. Calculate the acceleration of the bus between 7s - 10s.
Answers
Answer:
t=0, the particle is at rest say at the origin. After that the velocity is positive, so that the particle moves in the positive x direction. Its speed increases till 1 second when it starts decreasing. The particle continues to move further in positive x direction. At t=2s, its velocity is reduced to zero, it has moved through a maximum positive x distance. Then it changes its direction, velocity being negative, but increasig i magnitude.At t=3s velocity is maximum in the negative x direction and then the magnitude starts decreasing. it comes to rest at t=4 s.
a. Distance during 0 to 2 s = Area of OAB
=12×2s×10ms=10m=12×2s×10ms=10m
b. Distance during 2 to 4s = Area of BCD = 10 m. The partile has mioved in negative x direction during this period.
c. The distance travelled during 0 to 4s=10 m+10m
=20m=20m
d. displacement during 0 to 4 = 10 m+(-10m)=0
e. at t=1/2 s acceleration = slope of line OA=10 m/s^2
f. at t=2 s acceleration= slope of line ABC = - 10 m/s^2.