The diagram above shows a velocity time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation of the motion

VPETERS2002

6 years ago

Distance during acceleration
=area of triangle
=1/2*base*height
=1/2*10*10
=50m

Distance during retardation
=area of triangle
=1/2*base*height
=1/2*5*10
=25m

Total distance
=50m+25m
=75m....ans

olyezema

4 years ago

During the acceleration and deceleration periods of motion not the total distance covered
1/2(10+5)10
=75m

WissyJ

2 years ago

The question indicates the "total distance during acceleration and retardation " excluding uniform acceleration.
Therefore, 75m is correct.

olyezema

4 years ago

Pls this answer is wrong.......I'm going to post the solving as my comments......pls take a look at it

Finesse11

3 weeks ago

the answer is correct.
periodt.

usadara

2 years ago

please do correction, the correct answer is A.

Temmy49

7 years ago

How is it A?