Question

A car traveling initially at +7.9 m/s accelerates uniformly at the rate of +0.72 m/s2 for a distance of 265 m. What is its velocity at the end of the acceleration? Answer in units of m/s. 10.0 points What is its velocity after it accelerates for 114 m? Answer in units of m/s

Answer 1

1. 21.07 m/s

2. 15.05 m/s

Given:

initial speed of the car, u = 7.9 m/s

distance covered by the car, d = 265 m

acceleration, a = 0.72
`m/s^(2)`

Solution:

To calculate the velocity at the end of acceleration, we use the third eqn of motion:

`v^(2) = u^(2) + 2ad`

`v^(2) = 7.9^(2) + 2* 0.72* 265`

`v = \sqrt{7.9^(2) + 2* 0.72* 265} = 21.07 m/s`

Now,

Velocity after it accelerates for a distance for 114 m:

Here d = 114 m

Again, from third eqn of motion:

`v^(2) = u^(2) + 2ad`

`v = \sqrt{7.9^(2) + 2* 0.72* 114} = 15.05 m/s`

Answer 2

in first case velocity =21.07 m/sec in second case velocity =15.05 m/sec

Step-by-step explanation:

We have given initial velocity u = 7.9 m/sec

Acceleration
`a=0.72m/sec^2`

Distance S = 265 m

Now according to third law of motion
`v^2=u^2+2as` here v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance

So
`v^2=7.9^2+2* 0.72* 265=444.01`

v = 21.07 m/sec

In second case s =114 m

So
`v^2=7.9^2+2* 0.72* 114=226.57`

v =15.05 m/sec