
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi ​ =632mi/h=632mi/h( 1mi 1609m ​ )( 3600s 1h ​ )=282m/s (a) taking v xf ​ =v xi ​ +a x ​ t with v xf ​ =0 a x ​ = t v xf ​ −v xf ​ ​ = 1.40s 0−282m/s ​ =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f ​ −x i ​ = 2 1 ​ (v xi ​ +v xf ​ )t= 2 1 ​ (282m/s+0)(1.40s)=198m