Question

♥️hey mates..will you please help me??please give me solution and explanation..♥️
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Answer 1

Answer:

Let the cars meet at a point x from the first car after time t

For first car

X = VT = 20t

For second car distance travelled is given by

S = (1/2) * (at^2)

If they have to meet then at time t

X = 100+S

20t = 100 + 0.5(at^2)

0.5(at^2) - 20t + 100 = 0

at^2 - 40t + 200 = 0

If you solve for quadratic equation to have real solutions,

B^2 - 4AC >= 0

(-40^2) - 4(a)(200) >= 0

1600 - 800a >= 0

a <= 2m/s^2

Therefore max acceleration a = 2m/s^2