Question

a car accelerates from rest at a constant rate of 2 metre per second square for sometime then it starts at a constant rate of 4 m per second square and comes to rest it remains in motion for 6 seconds then the maximum speed attained and the total distance travelled is

Answer 1

We have spilt time interval into two parts. One for acceleration and one for retardation

Answer 2

Answer:The maximum speed is 4 m/s and the total distance traveled is 24 m.

Explanation:

Given that,

Acceleration $a =2 m/s^2$

Retardation $a = -4 m/s^2$

Total time t = 6 sec

For first case,

Using equation of motion

$v = u+at$

$v = 2t$

For second case,

initial velocity = final veloity of first case

$v = u+at$

$v= 2t-4(6-t)$

$v = -24+6t$

At rest, v = 0

So, $-24+6t = 0$

$t = 4 sec$

The total distance is

$s = s_{1}+s_{2}$......(I)

$s_{1} = u_{1}t+\dfrac{1}{2}at^2$

$s_{1} =0+\dfrac{1}{2}\times2\times4\times4$

$s_{1}= 16 m$

$s_{2}=u_{2}t+\dfrac{1}{2}at^2$

$s_{2}= 8\times2-\dfrac{1}{2}\times 4\times2\times2\times$

$s_{2}=  8m$

Put the value of $s_{1}$ and $s_{2}$ in equation (I)

$s = 16\ m+8\ m$

$s = 24\ m$

The total distance traveled is 24 m.

The maximum speed is

Maximum speed = total distance/total time

$v_{max}= \dfrac{24}{6}$

$v_{max}= 4 m/s$

Hence, The maximum speed is 4 m/s and the total distance traveled is 24 m.