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

Answer:

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.

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  = $\frac{AE}{OE}=\frac{30}{5}= 6$

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  = $\frac{BD}{CD}=\frac{30}{10}= 3$

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,

$A=(\frac{1}{2}\times OE\times AE)+ (\frac{1}{2}\times BD\times CD)+(AB\times BD)\\A=(\frac{1}{2}\times 5\times 30)+ (\frac{1}{2}\times 30\times 10)+(15\times30 )\\A=75+450+150=675 \textrm{ m}$.

Hence, total distance covered is 675 m.