Question

A particle under a constant acceleration is moving in a straight line and covers a distance of 20 m in first 2 sec and 40 m in next 5 sec . Calculate distance covered in subsequent 3 sec and total distance it covers before it comes to rest

Answer 1

distance covered in subsequent 60

total distance 80

Answer 2

Answer:

So, the distance covered in subsequent 3 seconds is $v = 34\frac{m}{s}$ and total distance is 156m.

Explanation:

Assume that the body's starting speed is u m/s and that the constant acceleration is a $\frac{m}{s^{2} }$.

The body's displacement (distance travelled) is then indicated after t seconds by

$s = ut + \frac{1}{2} at^{2}$ …. (1)

The body covers 20 m in 2s and 40 m in 5s.

$20 = 2u + \frac{1}{2} a (2)^{2} \\20 = 2u + 2a\\u + a = 10 .... (2)$

$40 = 5u + \frac{1}{2}a(5)^{2} \\40 = 5u + \frac{25}{2}a \\2u +5a = 16 ...(3)$

Solving for (2) and (3) for u and a we get

u + a = 10 ….(2) x 5

2u + 5a = 16 …(3) x 1

5u + 5a = 10 ….(4)

2u + 5a = 16  ….(5)

On subtracting (4) from (5), we get:

u = -2$\frac{m}{s^{2} }$

Put u = -2 $\frac{m}{s^{2} }$ in equation (2) we get:

a = 12 $\frac{m}{s^{2} }$

To obtain the distance covered in next 3 seconds we need to find the velocity beginning of this interval using,

v = u + at

v = -2 + 12(3)

v = 34 $\frac{m}{s}$

$s = ut +\frac{1}{2}at^{2}$

Substituting values we get:

s = 34 x 3 + $\frac{1}{2}$ x12 x $3^{2}$

s = 156 m

Hence, the body will cover 156 m in the next 3 seconds.

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