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Question

As a car passes the point A on a straight road, its speed is 10 m/s. The car moves with constant acceleration a m/s2 along the road for T seconds until it reaches the point B, where its speed is v m/s. The car travels at this speed for a further 10 seconds, when it reaches the point C. From C it travels for a further T seconds with constant acceleration 3a m/s2 until it reaches a speed of 20 m/s at the point D. Choose the correct option(s).
[Given that the distance between A and D is 675 m]

A
value of v is 12.5 m/s
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B
value of a is 18 m/s2
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C
value of T is 10 s
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D
value of T is 20 s
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Solution

The correct option is D value of T is 20 s
Let us try to understand the situation given in the question with the help of graphical representation.

From A to B, u=10 m/s, acceleration =a m/s2
Using first equation of motion, we can say
10+aT=v (1)
From C to D, acceleration =3a m/s2
Using first equation of motion, we can say
u+a2t=20 m/s
(10+aT)+3aT=20
10+4aT=20 (2)
From (1) and (2),
v=12.5 m/s .........(3)
Distance between A and D =675 m = Area under the curve
= Area under line AB + Area under line BC + Area under line CD
=12T(10+v)+10v+12(v+20)T=675 .....(4)
Using eq. (3) in eq. (4), we get
T=20 s.
From eq. (1), a=12.51020=18 m /s2
Hence, options (a), (b) and (d) are correct.

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