Question

A race car starting from rest accelerates at a constant rate of 5.00 m/s^2. (a) What is the velocity of the car after it has traveled 100 ft? (b) How much time has elapsed?

Answer 1

Answer:

GIVEN:

A CAR STARTING FROM REST STARTING AT A CONSTANT RATE OF 5 M/S^2.

TO FIND:

a.VELOCITY AFTER IT HAS TRAVELLED 100 FT.

b.TIME TAKEN TO REACH 100 FT.

FORMULAS TO BE USED:

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

$v = u + at$

SOLUTION:

INITIAL VELOCITY = 0 M/S

ACCELERATION =5 M/S^2

DISTANCE TRAVELLED =100 FT=30.5 m (1 ft= 0.305 m)

NOW FINAL VELOCITY:

=>V^2=U^2+2AS

=>V^2=0^2+2×5×30.5

=>V^2=0+10×305/10

=>V^2=305

=>V=

$\sqrt{305} m \: {s}^{ - 1}$

NOW TIME TAKEN:

$= > v = u + at \\ \\ = > \sqrt{305} = 0 + 5t \\ \\ = > \sqrt{305} = 5t \\ \\ = > t = \frac{ \sqrt{305} }{5} seconds$

HENCE,

FINAL VELOCITY =ROOT 305 M/S

TIME TAKEN = ROOT 305/5 SECONDS.

Answer 2

Given:

A race car starting from rest accelerates at a constant rate of 5.00 m/s².

To find:

• Velocity of car after travelling 100 ft ?
• Time elapsed ?

Calculation:

• 100 feet = 30.48 metres.

Applying EQUATIONS OF KINEMATICS:

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

$\rm \implies {v}^{2} = {0}^{2} + 2 \times 5 \times 30.48$

$\rm \implies {v}^{2} = 304.8$

$\rm \implies v = 17.45 \: m {s}^{ - 1}$

Now, time elapsed be 't':

$\rm \therefore \: t = \dfrac{v - u}{a}$

$\rm \implies \: t = \dfrac{17.45 - 0}{5}$

$\rm \implies \: t = 3.49 \: sec$

So, velocity is 17.45 m/s and time elapsed is 3.49 sec.