To solve this problem, we need to set the distances traveled by the car and the bus equal to each other and solve for time.

The equation for distance traveled under constant acceleration is d = 0.5*a*t^2, where d is distance, a is acceleration, and t is time.

For the bus, the equation is d_bus = 0.5*2*t^2 = t^2.

For the car, the equation is d_car = 200 + 0.5*4*t^2 = 200 + 2t^2.

Setting these equal to each other gives t^2 = 200 + 2t^2.

Solving for t gives t = sqrt(200) = 14.14 seconds.

So, the car will catch up with the bus after approximately 14.14 seconds.

Question

To solve this problem, we need to set the distances traveled by the car and the bus equal to each other and solve for time.

The equation for distance traveled under constant acceleration is d = 0.5*a*t^2, where d is distance, a is acceleration, and t is time.

For the bus, the equation is d_bus = 0.5*2*t^2 = t^2.

For the car, the equation is d_car = 200 + 0.5*4*t^2 = 200 + 2t^2.

Setting these equal to each other gives t^2 = 200 + 2t^2.

Solving for t gives t = sqrt(200) = 14.14 seconds.

So, the car will catch up with the bus after approximately 14.14 seconds.

Knowee AI · Accepted Answer

To solve this problem, we need to set the distances traveled by the car and the bus equal to each other and solve for time.

The equation for distance traveled under constant acceleration is d = 0.5*a*t^2, where d is distance, a is acceleration, and t is time.

For the bus, the equation is d_bus = 0.5*2*t^2 = t^2.

For the car, the equation is d_car = 200 + 0.5*4*t^2 = 200 + 2t^2.

Setting these equal to each other gives t^2 = 200 + 2t^2.

Solving for t gives t = sqrt(200) = 14.14 seconds.

So, the car will catch up with the bus after approximately 14.14 seconds.

A car is standing 200 m behind a bus, which is also at rest.The two start moving at the same instant but with differ-ent forward accelerations. The bus has acceleration 2 m/s2and the car has acceleration 4 m/s2. The car will catch upwith the bus after a time of