Question

A car P travelling at a constant speed of 20m/s overtakes another car Q which is moving at constant acceleration of 5ms^(-2) .Car Q starts from rest at the instant car P starts overtaking it.Assuming that the length of each car is 20m the total road distance used in overtaking in kilometers is sqrt(0.0k) .Then the value of (k+1)^(2) is​

Answer 1

Answer: The value of (k+1)² is 81.

Explanation:

To overtake Q, total distance which P has to cover = Distance covered by Q + Sum of length of the two cars.

Let P overtake Q in time t seconds

Distance covered by P = 20t metre

Distance covered by Q = $\frac{5t^{2} }{2}$

Therefore,

⇒ 20t =  $\frac{5t^{2} }{2}$ + 40

⇒ 40t =  ${5t^{2} }$ + 80

⇒ $5t^{2} -40t + 80 = 0$

⇒ t = 4

Total length used in overtaking = 20t = 80m = 0.08 km

0.0k = 0.08

k = 8

k+1 = 8+1 =9

(k+1)² = 9² = 81

#SPJ3

Answer 2

Answer:

4

Explanation:

20t=20+5/2t^2+20

=t^2-8t+16

=t=4

=-20+20t=20+20(4)

=100m

=0.1km

0.1=root of 0.0k

0.01=0.0k

k=1

(k+1)^2

(1+1)^2

=4