Question

A car moving with constant acceleration covers the distance between two points 180 m apart in 6 second its speed as it passes the second point is 45 M per second what is the acceleration and its speed at the first point?

Answer 1

V=45 m/s
u=?
a=?
S=180m
T=6 s
We know that
V=u + at
45=u+6a. . Equation 1
By second equation of motion
S=ut +1/2at^2
180=6u +18a equation 2
Solving one and 2 we get
U=15
A=5

Answer 2

The initial velocity is 15 m/s and acceleration is $\bold{5 \ m/s^2}$

Explanation:

We have been given the distance between two points (s) = 180 m  

Time taken to cover the distance (t) = 6 s

Final velocity (v) = 45 m/s

So we need to find initial velocity and acceleration.

The first equation of motion is,

$v=u+at$

So,

$45=u+6 a$

$u=45-6 a \rightarrow(1)$

From the second equation of motion:

$s=u t+\frac{1}{2} a t^{2} \rightarrow(2)$

Substitute (1) in (2),

$180=(45-6 a) 6+\left(\frac{1}{2} a \times 36\right)$

$180=270-36 a+\frac{36 a}{2}$

$180=270-36 a\left(1-\frac{1}{2}\right)=270-36 a\left(\frac{1}{2}\right)$

$180=270-18a$

$18 a=270-180$

$18a=90$

$\therefore a=\frac{90}{18}=5 \ m / s^{2}$

Substitute the acceleration value in equation (1), we get,

$u=45-(6 \times 5)$

$u=45-30=15 \ \mathrm{m} / \mathrm{s}$

Thus, the initial velocity is 15 m/s and acceleration is $5 \ m/s^2$