Question

study the velocity time graph and calculate
a)acceleration from a to b
b)acceleration from b to c
c)the distance covered in the region ABE and BCFE
d)average velocity from c to d

Answer 1

a) The initial​ speed of the car is the speed at t=0. Hence, it is 10 km/ h.

b) The maximum speed is 35 km/hr. See the points B and C.

c) B to C shows zero acceleration as there is no change in speed of the car with respect to time.

d) The part from C to D shows varying retardation as its speed is not decreasing uniformly with the time. (You can see that in the part A to B we have uniform acceleration.)

e)To calculate the distance travelled in the first 8 hrs, you need to calculate the area below AB and BC, which is,

(35+10)×3/2 + 35× 5
67.5 + 175 = 242.5 km

You can also calculate this by applying the equations of motion.
Let me calculate the distance from A to B using the equations of motion.
u= 10
v=35
t= 3
a =( v-u)/t = (35-10)/3= 25/3
v²= u² +2as
s = (v² - u²)/ 2a
putting all​ the values we get, s= 67.5 km, which is equal yo the area of trapezium formed below AB. Hence, total distance = 67.5+175=242.5 km. 

I think I have answered all of them correctly.

Answer 2

Answer:

hey mate!!

Explanation:

1)8.3$m/s^{2}$

2)-11$m/s^{2}$

3)75m

4)7.5m/s

5)20m