Question

A car is accelerated 5.5 ft/s^2. Calculate the initial velocity v, the car must have if it is to attain a final velocity v of 45 mph in a distance of 352 ft. Compute also the time t required to attain that final velocity.

Answer 1

Answer:

1) The initial velocity must be 22 feet/sec.

2) The time required to change the velocity is 8 seconds.

Explanation:

This problem can be solved using third equation of kinematics

According to the third equation of kinematics we have

$v^{2}=u^{2}+2as$

where

'v' is the final velocity of object

'u' is the initial velocity of object

'a' is acceleration of the object

's' is the distance in which this velocity change is brought

Since the final velocity is 45 mph converting this to feet per seconds we get

45 mph =66 feet/sec

Applying Values in the above equation and solving for 'u' we get

$66^{2}=u^{2}+2\times 5.5\times 352\\\\u^{2}=484\\\\\therefore u=22feet/sec$

2)

The time required to attain this velocity can be found using first equation of kinematics as

$v=u+at$

where

't' is the time in which the velocity change occurs

'v', 'u', 'a' have the usual meaning

Thus applying given values we get

$66=22+5.5\times t\\\\\therefore t=\frac{66-22}{5.5}=8seconds$

Applying the given values we get