Atruck driver has a shipment of apples to deliver to a destination 550 miles away. the trip usually takes him 10.0 hours. today he finds himself daydreaming and realizes 120 miles into his trip that that he is running 30.0 minutes later than his usual pace at this point. if the driver still wishes to complete the trip in 10.0 hours, how fast must he drive for the rest of the trip? (in all questions, you may assume that the truck moves with a constant speed.)

at what speed must he drive for the remainder of the trip to complete the trip in the usual amount of time? express your answer using three significant figures.

He must drive at 58.8mi/hr

{3 significant figure}

Explanation: The original speed of the car to complete the journey normally is = 550mile/10hr

= 55mi/hr

The normal time to cover 120 mile with that speed is

t = distance/speed

= 120/55

=2hr, 0.18*60=11mins

= 2hr11mins.

But the driver found out he was 30mins behind at this distance, so he spent

2hr 11min +30min= 2hr41mins

For him to meet up the original schedule time 10hrs, he has

(10-2.41)hrs = 7hr19mins to cover the remaining distance of

{550-120}=430mile.

First, 7hr19mins= {7*(19/60)}hr

= {439/60}hr

Now let's find the speed at which the driver must move for him to cover for his initial delay

Speed= distance/time

= 430/{439/60}

This is same as writing

Speed = 430/439/60

Using the law of reciprocal,

Speed = (430*60)/439

= 25800/439

= 58.7699mi/hr

But we were asked to leave our answer in 3 significant figure. Therefore,

Speed = 58.8mi/hr

the force required to move the cart is divided by three or 33.33 n.