Answer: Given, the car is initially at rest; initial velocity (u) = 0 ms-1 Acceleration (a) = 4 ms-2 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 (4ms-2)(10s)2 = 200 meters Therefore, the car will cover a distance of 200 meters after 10 seconds. 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 Is there an error in this question or solution? 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
Open in App Suggest Corrections |