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a car start from rest and accelerate uniformly until it reaches a velocity of 30m/s...

a car start from rest and accelerate uniformly until it reaches a velocity of 30m/s after 5 seconds. it travels with uniformly velocity for 15 seconds and it then brought to rest in 10 seconds with the uniform retardation. calculate the total distance travel;
1.after the first 5 seconds
2.after 30 seconds?

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Answers (2)

PaulB
1 week ago
If you draw the velocity time graph you will have a trapezium having two two right angle triangle and a rectangle.
1) u=0
T=5
a=30/5=6m/s²
S=ut +½at²
S=0 +½*6*5²
S= 75metres in the first 5 seconds

2) distance in 30 sec
Since we have the distance for the first 5secs it then leaves us with 25 secs

When the body was in uniform velocity of 3om/s for 15 secs. The distance covered will be the area of the shape which in this case is a rectangle.
Area or rectangle = L X B=30 x 15=450metres

For the deceleration part for 10 secs at 30m/s
The shape here is a triangle.
Area of triangle =½L xB =10 x 30/2. =75metres

Therefore distance in 30 secs is 75m +450m +75m =675m
eze
3 weeks ago
(1) total distance after 5s is 75m
(2) total distance after 30s is 675m
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