Question

a car travels from rest and accelerate uniformly for 5seconds untill it attains a velocity of 30seconds it then travels with uniform velocity for 15seconds before decelerating uniformly to rest i 10seconds sketch a graph of the motion . using the graph above calculate the acceleration during the first 5seconds . deceleration during the last 10seconds .. total distance covered throughout the motion​

Answer 1

Acceleration during first 5 seconds is 6 m/s².

Deceleration during last 10 seconds is 3 m/s².

Total distance covered throughout the motion is 675 m.

Step-by-step explanation:

The sketch showing the variation of velocity with time is given below.

Slope of the line OA gives the acceleration during first 5 seconds.

∴ Slope of line OA =
`(AE)/(OE)=(30)/(5)= 6`

Therefore, acceleration during first 5 seconds is 6 m/s².

Now, absolute value of slope of line BC is the deceleration during last 10 seconds.

∴ Slope of line BC =
`(BD)/(CD)=(30)/(10)= 3`

Therefore, deceleration during last 10 seconds is 3 m/s².

Now, area under the graph gives total distance covered.

Area under the graph is given as the sum of areas of triangle OAE, triangle BCD and rectangle BCDE.

∴Area under the graph is,

`A=((1)/(2)* OE* AE)+ ((1)/(2)* BD* CD)+(AB* BD)\\A=((1)/(2)* 5* 30)+ ((1)/(2)* 30* 10)+(15*30 )\\A=75+450+150=675 \textrm{ m}`.

Hence, total distance covered is 675 m.