A racing car has a uniform acceleration of 4 ms^(-2 what distance will it cover in 10 s after start)

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The correct answer is 200 m.

• Given that, the car is initially at rest
  • Hence, initial velocity (u) = 0 m/s
  • Acceleration (a) = 4 m/s2
  • Time period (t) = 10 s
  • As per the second motion equation, s = ut + 1/2 at2

Therefore,

• The total distance covered by the car (s) = 0 × 10m + 1/2 (4m/s2)(10s)2
• = 200 meters
• Therefore, the car will cover a distance of 200 meters after 10 seconds.

• There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a).
• The following are the three equation of motion:
  • First Equation of Motion: v = u + at
  • Second Equation of Motion: s = ut + 1/2(at2)
  • Third Equation of Motion: v2 = u2 + 2as

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A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?

Initial velocity of the racing car, u = 0 (since the racing car is initially at rest)

Acceleration, a = 4 m/s2

Time taken, t = 10 s

According to the second equation of motion:

`s=ut+1/2at^2`

Where, s is the distance covered by the racing car

`s=0+1/2xx4xx(10)^2=400/2=200m`

Hence, the distance covered by the racing car after 10 s from start is 200 m.

Concept: Equations of Motion by Graphical Method - Derivation of Velocity - Time Relation by Graphical Method

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Text Solution

200 m500 m900 m400 m

Answer : A

Solution : Here, acceleration, `a = 4 ms^(-2)`, time taken, `t = 10 s` <br> initial velocity, `u = 0`, distance, `s = ?` <br> From `s = ut + (1)/(2)at^(2)`, `s = 0 xx 10 + (1)/(2) xx 4 (10)^(2) = 200 m`

Text Solution

Solution : Initial Velocity of the car, u = 0`ms^(-1)` <br> Acceleration, a = 4m `s^(-2)` <br> Time, t = 10 s <br> We know Distance, s = ut + (1/2) `at^(2)` <br> Therefore, Distance covered by car in 10 second = `0xx10+(1//2)xx4xx102` <br> = `0+(1//2)xx4xx10xx10m ` <br> = `(1//2) xx400m ` <br> = 200m