Question

1) diagram shows velocity-time graph A and B starting from the same place and moving along a road in the same direction calculate.
i) acceleration of car A
ii) acceleration of car B between 2 seconds and 4 seconds
iii) at wgat time interval do both cars have same velocity
iv) which if the 2 cars is ahead and by how much after 8 seconds​

Answer 1

Answer:

(i) Acceleration of car A =  10$m/s^{2}$

Explanation:

Acceleration (a) = $\frac{v-u}{t}$ = $\frac{10-1}{1}$ = $10 m/s^{2}$

(ii) Acceleration of the car B between 2 sec and 4 sec = 20 $m/s^{2}$

Explanation:

Acceleration (a) = $\frac{v-u}{t}$ = $\frac{60-20}{2}$ = $\frac{40}{2}$ = $20 m/s^{2}$.

(iii) Time interval when both cars have same velocity = 6 sec.

Explanation:

From the graph it is clear that  both the lines are meeting at a point where the time is 6 sec.

Hence we can say that both the cars have same velocity at 6 sec.

(iv) After 8 sec,  car A is 160 m ahead of car B.

Explanation:

Displacement (s) = velocity X time.

After 8 sec,

for car A , velocity = $80 m/s^{2}$

         Therefore, s = 80 X 8 = 640 m.

for car B , velocity = $60 m/ s^{2}$

Therefore,           s = 60 X 8 = 480 m.

Difference in their displacements = 640 - 480 = 160 m.