In the daytona 500 auto race, a ford thunderbird and a mercedes benz are moving side by side down a straightaway at 71.5 m/s. the driver of the thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of 250 m. she spends 5.00 s in the pit and then accelerates out, reaching her previous speed of 71.5 m/s after a distance of 350 m. at this point, how far has the thunderbird fallen behind the mercedes benz, which has continued at a constant speed?

957.4845 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Time taken by the thunderbird to stop is 6.993 seconds

Time the thunderbird was at the pit is 5 seconds

Time taken to accelerate back to 71.5 m/s is 9.79 seconds

Total time to this point is 6.993+5+9.79 = 21.783 seconds

The Mercedes Benz is moving at a constant velocity hence it has no acceleration and we use the formula

Distance = Speed × Time

⇒Distance = 71.5 × 21.783 = 1557.4845 m

The thunderbird has covered 250+350 = 600 m

So, the distance between them is 1557.4845-600 = 957.4845 m

the correct choice is a .

do not post any more test or exam questions on

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach