Question

A particle starts from rest at a fixed point A and moves in a straight line with an acceleration which, t seconds after leaving A, is given by a = 4t. After 2 seconds the particle reaches a point B and the acceleration then ceases. Find:
i) the velocity when the particle reaches B
ii) the distance AB

The particle moves on immediately with acceleration given by -3t, where t seconds is the time after the particle leaves A, until it comes to rest at a point C. Find:
iii) the value of t when the particle reaches C
(iv) the distance AC​

Answer 1

(a) The velocity when the particle reaches B is 8 m/s.

(b) The distance between point A and B is 5.33 m.

(c) The value of t when the particle reaches C is 1.63 seconds.

(d) The distance AC​ is 8.7 m.

Velocity when the particle reaches B

The velocity when the particle reaches B is calculated as follows;

v = ∫a. dt

where;

• v is velocity
• a is acceleration of the particle

v =  ∫(4t . dt)

v = 4t²/2

v = 2t²

v(2) = 2(2)²

v(2) = 8 m/s

Thus, the velocity when the particle reaches B is 8 m/s.

Distance AB

x = ∫v

where;

• x is the distance between  A and B

x = ∫2t². dx

x = ²/₃t³

x(2) = ²/₃(2)³

x(2) = 5.33 m

Thus, the distance between point A and B is 5.33 m.

value of t when the particle reaches C

when the particle reaches point C, final velocity, vf = 0

vf = v + at

where;

• v is the velocity at point

0 = 8 - 3t(t)

0 = 8 - 3t²

3t² = 8

t² = 8/3

t² = 2.67

t = √(2.67)

t = 1.63 seconds

Distance between A and C

x = ∫vf

x = ∫(8 - 3t²)

x = 8t  - t³

x(1.63) = 8(1.63) - (1.63)³

x(1.63) = 8.7 m

Learn more about velocity here: brainly.com/question/24681896

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