Question #3086

A motor car starts moving at A and moves to B by uniform acceleration.It moves B to C by uniform velocity that has a value of 10m/s and moves C to D by uniform retardation and stops.It takes 35s to move A to D and BC=200m.Draw a velocity time graph for the motion of car.(the motor car starts with motionless and it moves in a straight way)

1.what is the car's acceleration?

2.what time period it moves by acceleration?

3.mention the distance that the car moves by acceleration.

4.mention the distance that the car moves by retardation.

1.what is the car's acceleration?

2.what time period it moves by acceleration?

3.mention the distance that the car moves by acceleration.

4.mention the distance that the car moves by retardation.

Expert's answer

First we should find out by what time the car moved with acceleration

Let t1 - time on AB

t2-time on BC

t3-time on CD

t2=200/10=20s

t1=t3=(35-20)/2=7.5s - time period when car moves with acceleration

on AB v=a*t, a- acceleration, v-velocity

Hence a=v/t

a=10/7.5=1.33 m/s^{2}

The distance that car moves by acceleration s= a*t^{2}/2

t=7.5

a=1.33

Hence s=1.33*7.5^{2}/2=37.5 m

The distance that car moves by retardation is the same to the distanse by acceleration

While solving this task we assume that the absolute value of the acceleration on AB is similar to the absolute value of the retardation on CD. If it isn't so the problem hasn't got a unique solution

Let t1 - time on AB

t2-time on BC

t3-time on CD

t2=200/10=20s

t1=t3=(35-20)/2=7.5s - time period when car moves with acceleration

on AB v=a*t, a- acceleration, v-velocity

Hence a=v/t

a=10/7.5=1.33 m/s

The distance that car moves by acceleration s= a*t

t=7.5

a=1.33

Hence s=1.33*7.5

The distance that car moves by retardation is the same to the distanse by acceleration

While solving this task we assume that the absolute value of the acceleration on AB is similar to the absolute value of the retardation on CD. If it isn't so the problem hasn't got a unique solution

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