Question

A body starting from rest and travelling with uniform acceleration has a velocity of 40m/s after 10 second at A. Velocity of the body 4 second before it crosses the point A is
a)16m/s
b)20m/s
c)24 m/s
d)32m/s

Answer 1

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0, v = 40m/s, t = 10s and we want to find acceleration, a, so we use equation (3) v = u + at

40 = 0 + 10a

So a = 4m/s^2

Now we have u = 0, t = 6s, a = 4m/s^2 and we want to find v, so we use equation (3) v = u + at

v = 0 + 4(6) = 24m/s

So the velocity at 6s is 24m/s

NOTE: there are quicker ways to solve this problem, for example we could just calculate v = 6/10 x 40 = 24m/s, but I have solved it “the long way” to demonstrate the method.


HOPE IT HELPS

Answer 2

Let the
Initial velocity be (u) = 0m/s
Final velocity be (v) = 40m/s
Time taken to reach (v) be (t) = 10s

As we know that,
$\: \: \: \: \: \: \: \: \: \: v = u + at \\ = > 40 = 0 + 10a \\ = > a = \frac{40}{10} \\ = > a = 4 \frac{m}{ {s}^{2} }$


4 seconds before it crosses A
Means
6 seconds after the body has started.
Acceleration remains same here.
So,
Again,
$\: \: \: \: \: \: v = u + at \\ = > v = 0 + 4 \times 6 \\ = > v = 24 \frac{m}{ {s}}$

Therefore,
Required answer is
24m/s.



Hope this will help you. 

If you like my answer
Please mark it as brainliest 
And 
Be my follower if possible.