Question

Hey guys
Plss helo me answr thus and make ke understand this ques of class 9

Answer 1

it's d..............

Answer 2

Part a

Answer: Car A moves ahead.

We are given the distance of A and B from the starting point as:$x_A(t)=\alpha t+\beta t^2$

$x_B(t)=\gamma t^2 -\delta t^3$

where,

$\alpha =2.6 m/s\\ \beta =1.20 m/s^2 \\ \gamma = 2.80 m/s^2\\ \delta=0.20m/s^3$

Just after leaving from the starting point,

say t = 1 s

$x_A(1)=2.6 m/s \times (1s)+1.20m/s^2(1s)^2 = 3.8 m$

$x_B(1)= 2.80m/s^2(1s)^2 - 0.20 m/s^3(1s)^3=2.6 m$

Since $x_A>x_B$ at t =1 s, Car A is ahead of car B. Hence, option A is correct.

Part B

Answer: The cars will have same position after 2.6 s and 5.73 s after leaving from starting position.

Explanation:

We need to find the time at which the two cars are at same position. So apart from starting point at t =0, the cars would be at same position when

$x_A(t)=\alpha t+\beta t^2=x_B(t)=\gamma t^2 -\delta t^3$

$\Rightarrow \alpha t+\beta t^2=\gamma t^2 -\delta t^3$

Substitute the values and solve for t

$\Rightarrow 0.20 t^3-1.6t^2+2.6t=0\\ \Rightarrow t(0.20t^2-1.6t+2.6)=0$

t=0 ( starting position)

Solve for the quadratic equation:

$(0.20t^2-1.6t+2.6)=0$

$\Rightarrow t=\frac {1.6\pm\sqrt{1.6^2-4\times2.6\times 0.20}}{2\times0.2}$

Solving this gives two values, t = 2.6 s and 5.73 s

Hence, the correct option is (d). The cars will have same position after 2.6 s and 5.73 s after leaving from starting position.