Question
A motorbike initially travelling at a velocity of 4m/s accelerates to a velocity of 12m/s The motorbike travels a distance of 16 m during this time. Calculate the acceleration of the motorbike.
Answer
4
(280 Votes)
Sheldon
Elite · Tutor for 8 years
Answer
The proper acceleration of the motorbike is a = 4 m/s^2.
Explanation
Acceleration can be calculated using the formula a = (v_f - v_i)/t, where a = acceleration, v_f = final velocity, v_i = initial velocity, and t = time. However, in this problem, we are not given the time directly. Instead, we first need to calculate the time using the formula for distance (s) in uniformly accelerated motion: s = v_it + 0.5a*t^2. This problem requires us to solve first for time and then use that result to calculate acceleration.In this case:Initial velocity (v_i) = 4 m/s,Final velocity (v_f) = 12 m/s,and the total distance (s) = 16 m.Initial step involves rearranging the formula for distance to solve for time.'Given, s = v_it + 0.5a*t^2 => t = (s - 0.5vat)/v_i (equation 1)But we don't have acceleration and the equation includes the term of 'time(t)', we need to bring the other equation: v_f = v_i + t*a (equation 2).From equation 2: t = (v_f - v_i)/a (equation 3), and now substitute EQ.3 into EQ.1, we could get the relation of acceleration (a) only and then find the specific value.Calculation:Rearrange equation 2 to get:a = (v_f - v_i)/t,Now substitute this into equation 1 to get acceleration:=> A = [(v_f - v_i)^2 / (2 * (s - (v_i*(v_f - v_i)/ (v_f - v_i()))))] ---- Apply the value into equation becomes = [(12 - 4)^2 / (2*(16 - 4*(12 - 4)/(12 - 4)))] = (8^2 / (2*(16 - 8)) = 64 / 16 = 4 m/s^2