Question

a lorry moving under uniform acceleration covers a distance of 18 m from point a to point B along a straight line path in 15 seconds its velocity while passing through the point B is 2 metre per second what is the velocity of the Lori at the point A what is it's acceleration

Answer 1

Final Answer :
Velocity at point A ,u = 0.4 m/s
acceleration ,a = 0.11 m/s^2

Steps and Understanding :

1)
Initial Velocity at point A , = u (say)
Final Velocity at point B, v = 2m/s
Time taken to move from A to B = 15s
acceleration = a (say)


2) By Applying Newtons Equation of motion.

S = ut + 1/2 at^2
=>
$18 = u \times 15 + \frac{1}{2} \times a \times {15}^{2} \\ = > 18 = 15 u + \frac{225a}{2} - - - - (1)$


3) And,
v = u + at
=> 2 = u + a*15
$= > 2 = u + 15a \: - - - (2)$


4) Using equation (1) & (2) ,
$= > 18 = 15u + \frac{15(2 - u)}{2} \\ = > 18 = 15u + 15 - \frac{15u}{2} \\ = >3 = \frac{15u}{2} \\ = > u = \: \: 0.4m \div s$
5) Then,
$a = \frac{(2 - u)}{15} = \frac{2 - 0.4}{15} = 0.11 \frac{m}{ {s}^{2} }$

Hence,
Velocity at point A = 0.4m/s
Acceleration = 0.11 m/s^2

Answer 2

Answer:4m/s

Explanation:s=1/2(u+v)t

s=180 m

v=20m/s

t=15 s

Putting values in formula we get:

180=1/2(u+20)x15

180x2=(u+20)15

360/15=u+20

u+20=24

u=24-20

u=4m/s

Therefore its point while passing through point A is 4m/s