The correct option is D Acceleration is 4 ms−2at t=0
Average velocity →v=∫50vdt∫50dt=∫50(4t−t2)dt∫50dt
=[2t2−t33]505=50−12535=253×5=53 ms−1
For average speed,
let us put v=0, which gives t=0 and t=4 ie velocity changes sign at t=4.
∴ Average speed = ∣∣∣∫40vdt∣∣∣+∣∣∣∫54vdt∣∣∣∫50dt
=∣∣∣∫40(4t−t2)dt∣∣∣+∣∣∣∫54vdt∣∣∣5
=[2t2−t33]40+[−2t2+t33]545
=(32−643)+[−2(25−16)+13(125−64)]5=135 ms−1
For acceleration:
a=dvdt=ddt(4t−t2)=4−2t
At t=0,a=4 ms−2
∴ Options c and d are correct