Motion and Constant Acceleration. A speeding car is traveling at a constant speed of $30.0m/s$ when it passes a stopped police car. The police car accelerates at $7.0m/s^2$ . How fast will it be going when it catches up with the speeding car?

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Answer 1 · 2018-02-24T15:36:09

The speed of the police car is $=60ms^-1$

To catch the speeding car, the distance travelled is the same by both cars,

For the speeding car,

$d=ut=(30t )"m"$

For the police car,

The initial speed is $u=0$

$d=ut+1/2at^2=0+1/2*7*t^2$

Therefore,

$30t=7/2t^2$

$30t-7/2t^2=0$

$t(30-7/2t)=0$

$t=0$ which is the initial condition

$30-7/2t=0$ , $=>$ , $t=60/7s$

Therefore,

The speed of the police car is

$v=u+at=0+7*60/7=60ms^-1$

Motion and Constant Acceleration. A speeding car is traveling at a constant speed of 30.0m/s30.0m/s when it passes a stopped police car. The police car accelerates at 7.0m/s^27.0m/s^2. How fast will it be going when it catches up with the speeding car?